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2160 Compendium
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MATH 2160 — Question Compendium
Every question from the 23 completed MyLab assignments (Lay, Linear Algebra and Its Applications, 6e), transcribed to LaTeX with the recorded answer. Linked from the chapter study guides.
343 questions across 22 sections.
Extracted from MyLab review pages. Math came from Pearson’s accessible verbalisation and @MATX encodings rather than OCR, so matrices are exact. Answers are the ones recorded on your attempts.
§1.1 Systems of Linear Equations
1.1.1
Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure. $x_{1} + 5 x_{2} = 8 2 x_{1} + 7 x_{2} = 7$
Answer
Find the solution to the system of equations. $(x_{1}, x_{2}) =$ (-7, 3) (Simplify your answer. Type an ordered pair.)
1.1.2
Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure. $7 x_{1} + 14 x_{2} = 28 3 x_{1} + 7 x_{2} = 9$
Answer
Find the solution to the system of equations. $(x_{1}, x_{2}) = (10, -3)$ (Simplify your answer. Type an ordered pair.)
1.1.3
Find the point $(x_{1}, x_{2})$ that lies on the line $x_{1} + 3 x_{2} = 10$ and on the line $x_{1} - x_{2} = 2$. See the figure. $x_{1} x_{2}$ A graph has a horizontal axis labeled $x_{1}$ and a vertical axis labeled $x_{2}$. A line on the graph falls from left to right and another line rises from left to right. The lines intersect in the first quadrant. The lines are not labeled and the axes do not show any scale
Answer
The point $(x_{1}, x_{2})$ that lies on the line $x_{1} + 3 x_{2} = 10$ and on the line $x_{1} - x_{2} = 2$ is (4, 2). (Simplify your answer. Type an ordered pair.)
1.1.6
Consider the accompanying matrix as the augmented matrix of a linear system. State in words the
Answer
D. Replace row 4 by its sum with -3 $\times$ row 3. (Type an integer or a simplified fraction.)
1.1.7
The augmented matrix of a linear system has been reduced by row operations to the form shown. Continue the appropriate row operations and describe the solution set of the original system.$\begin{bmatrix} 1 & 8 & 4 & -3 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & -1 & 2 \end{bmatrix}$
Answer
C. The solution set is empty.
1.1.10
The augmented matrix of a linear system has been reduced by row operations to the form shown. Continue the appropriate row operations and describe the solution set of the original system.$\begin{bmatrix} 1 & -2 & 0 & 0 & -4 \\ 0 & 1 & -1 & 0 & -5 \\ 0 & 0 & 1 & -3 & 2 \\ 0 & 0 & 0 & 1 & 4 \end{bmatrix}$
Answer
A. The solution set contains one solution: (14, 9, 14, 4). (Type integers or simplified fractions.)
1.1.13
Solve the system. $x_{1} - 6 x_{3} = 17 4 x_{1} + 2 x_{2} - 9 x_{3} = 44 x_{2} + 5 x_{3} = -7$
Answer
A. The unique solution of the system is $(x_{1}, x_{2}, x_{3}) =$ (5, 3, -2). (Type integers or simplified fractions.)
1.1.22
Do the three planes $x_{1} + 3 x_{2} + x_{3} = 5, x_{2} - x_{3} = 1$, and $x_{1} + 4 x_{2} = 4$ have at least one common point of intersection? Explain
Answer
A. The three planes do not have a common point of intersection.
1.1.27
Determine whether the statement below is true or false. Justify the answer. Every elementary row operation is reversible
Answer
C. The statement is true. Replacement, interchanging, and scaling are all reversible.
1.1.28
Determine whether the statement below is true or false. Justify the answer. Elementary row operations on an augmented matrix never change the solution set of the associated linear system
Answer
C. The statement is true. Each elementary row operation replaces a system with an equivalent system.
1.1.29
Determine whether the statement below is true or false. Justify the answer. $A_{5} \times 6$ matrix has six rows
Answer
D. The statement is false. $A_{5} \times$ ×6 matrix has five rows and six columns.
1.1.30
Determine whether the statement below is true or false. Justify the answer. Two matrices are row equivalent if they have the same number of rows
Answer
A. The statement is false. Two matrices are row equivalent if there exists a sequence of elementary row operations that transforms one matrix into the other.
1.1.31
Determine whether the statement below is true or false. Justify the answer. The solution set of a linear system involving variables $x_{1}$,…, $x_{n}$ is a list of numbers $(s_{1}$,…, $s_{n})$ that makes each equation in the system a true statement when the values $s_{1}$,…, $s_{n}$ are substituted for $x_{1}$,…, $x_{n}$, respectively
Answer
D. The statement is false. The given description is of a single solution of such a system. The solution set of the system consists of all possible solutions.
1.1.39
Find the elementary row operation that transforms the first matrix into the second, and then find the reverse operation that transforms the second matrix into the first.$\begin{bmatrix} -1 & 6 & -1 \\ 2 & -2 & 1 \\ -4 & 6 & -3 \end{bmatrix}$,$\begin{bmatrix} -4 & 6 & -3 \\ 2 & -2 & 1 \\ -1 & 6 & -1 \end{bmatrix}$
Answer
A. Interchange row 1 and row 3. Interchange row 1 and row 3. Interchange row 1 and row 3.
1.1.41
Find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.$\begin{bmatrix} 1 & -4 & 2 & 0 \\ 2 & -2 & 3 & -6 \\ 0 & 4 & -3 & 5 \end{bmatrix}$,$\begin{bmatrix} 1 & -4 & 2 & 0 \\ 0 & 6 & -1 & -6 \\ 0 & 4 & -3 & 5 \end{bmatrix}$
Answer
A. Replace row 2 by its sum with -2 $\times$ row 1. (Type an integer or a simplified fraction.)
§1.2 Row Reduction and Echelon Forms
1.2.2
Determine which matrices are in reduced echelon form and which others are only in echelon form. a.$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$b.$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$c.$\begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
Answer
Is matrix a in reduced echelon form, echelon form only, or neither? Your answer is correct. Part 2 Is matrix b in reduced echelon form, echelon form only, or neither? Your answer is correct. Part 3 Is matrix c in reduced echelon form, echelon form only, or neither? Your answer is correct
1.2.3
Row reduce the matrix to reduced echelon form. Identify the pivot positions in the final matrix and in the original matrix, and list the pivot columns.$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 7 \\ 7 & 8 & 9 & 10 \end{bmatrix}$
Answer
B. $\begin{bmatrix} \mathbf{1} & 0 & -1 & -2 \\ 0 & \mathbf{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
1.2.4
Row reduce the matrix to reduced echelon form. Identify the pivot positions in the final matrix and in the original matrix, and list the pivot columns.$\begin{bmatrix} 1 & 2 & 4 & 1 \\ 2 & 4 & 5 & -1 \\ 4 & 5 & 4 & -2 \end{bmatrix}$
Answer
A. $\begin{bmatrix} \mathbf{1} & 0 & 0 & 1 \\ 0 & \mathbf{1} & 0 & -2 \\ 0 & 0 & \mathbf{1} & 1 \end{bmatrix}$
1.2.8
Find the general solution of the system whose augmented matrix is given below.$\begin{bmatrix} 1 & 3 & 0 & 10 \\ 2 & 5 & 0 & 15 \end{bmatrix}$
Answer
C. $\{ \begin{bmatrix} x_{1} = -5 \\ x_{2} = 5 \\ x_{3} is free \end{bmatrix}$
1.2.9
Find the general solution of the system whose augmented matrix is given below.$\begin{bmatrix} 0 & 1 & -2 & 3 \\ 1 & -3 & 1 & -3 \end{bmatrix}$
Answer
A. $\{ \begin{bmatrix} x_{1} = 6 + 5 x_{3} \\ x_{2} = 3 + 2 x_{3} \\ x_{3} is free \end{bmatrix}$
1.2.11
Find the general solution of the system whose augmented matrix is given below.$\begin{bmatrix} 2 & -3 & 5 & 0 \\ 8 & -12 & 20 & 0 \\ 6 & -9 & 15 & 0 \end{bmatrix}$
Answer
B. $\{ \begin{bmatrix} x_{1} = three halves x_{2} - five halves x_{3} \\ x_{2} is free \\ x_{3} is free \end{bmatrix}$x1= 3 2 x2- 5 2 x3 x2 is free x3 is free
1.2.13
Find the general solution of the system whose augmented matrix is given below.$\begin{bmatrix} 1 & -6 & 0 & -1 & 0 & -8 \\ 0 & 1 & 0 & 0 & -5 & 4 \\ 0 & 0 & 0 & 1 & 8 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
A. $\{ \begin{bmatrix} x_{1} = \square \\ x_{2} = \square \\ x_{3} is free \\ x_{4} = \square \\ x_{5} is free \end{bmatrix} \{$
1.2.14
Find the general solution of the system whose augmented matrix is given below.$\begin{bmatrix} 1 & 0 & -4 & 0 & -7 & 2 \\ 0 & 1 & 5 & -1 & 0 & 6 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
C. $\{ \begin{bmatrix} x_{1} = \square \\ x_{2} = \square \\ x_{3} is free \\ x_{4} is free \\ x_{5} = \square \end{bmatrix} \{$
1.2.25
Determine whether the statement below is true or false. Justify the answer. In some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations
Answer
A. The statement is false. Each matrix is row equivalent to one and only one reduced echelon matrix.
1.2.26
Determine whether the statement below is true or false. Justify the answer. The echelon form of a matrix is unique
Answer
B. The statement is false. The echelon form of a matrix is not unique, but the reduced echelon form is unique.
1.2.34
Determine whether the statement below is true or false. Justify the answer. A general solution of a system is an explicit description of all solutions of the system
Answer
D. The statement is true. The row reduction algorithm leads directly to an explicit description of the solution set of a linear system when the algorithm is applied to the augmented matrix of the system, leading to a general solution of a system.
1.2.35
Suppose $a_{5} \times 7$ coefficient matrix for a system has five pivot columns. Is the system consistent? Why or why not?
Answer
C. There is a pivot position in each row of the coefficient matrix. The augmented matrix will have eight eight columns and will not have a row of the form$\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ so the system is consistent.
1.2.36
Suppose a system of linear equations has $a_{3} \times 5$ augmented matrix whose fifth column is not a pivot column. Is the system consistent? Why or why not?
Answer
To determine if the linear system is consistent, use the portion of the Existence and Uniqueness Theorem, shown below. A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column. That is, if and only if an echelon form of the augmented matrix has no row of the form$\begin{bmatrix} 0 & \cdots & 0 & b \end{bmatrix}$with b nonzero. Part 2 In the augmented matrix described above, is the rightmost column a pivot column? Your answer is correct. Part 3 In the echelon form of the augmented matrix, is there a row of the form$\begin{bmatrix} 0 & 0 & 0 & 0 & b \end{bmatrix}$with b nonzero? Your answer is correct. Part 4 Therefore, by the Existence and Uniqueness Theorem, the linear system is consistent
1.1
Solve the system of equations. $2 x_{1} + x_{2} = 0 x_{1} - 3 x_{2} + x_{3} = 0 3 x_{1} + x_{2} - x_{3} = 0$
Answer
C. (0, 0, 0)
§1.3 Vector Equations
1.3.1
Compute$\mathbf{u}$+$\mathbf{v}$and$\mathbf{u} - 4 \mathbf{v}$.$\mathbf{u}$=$\begin{bmatrix} -6 \\ 4 \end{bmatrix}$,$\mathbf{v}$=$\begin{bmatrix} -3 \\ 3 \end{bmatrix}$
Answer
$\mathbf{u}$+$\mathbf{v}$=$\begin{bmatrix} -9 \\ 7 \end{bmatrix}$(Simplify your answer.) Part 2$\mathbf{u} - 4 \mathbf{v}$=$\begin{bmatrix} 6 \\ -8 \end{bmatrix}$(Simplify your answer.)
1.3.5
Write a system of equations that is equivalent to the given vector equation. $x_{1} \begin{bmatrix} 5 \\ -5 \\ 9 \end{bmatrix} + x_{2} \begin{bmatrix} 4 \\ 0 \\ -9 \end{bmatrix}$=$\begin{bmatrix} 4 \\ -2 \\ 6 \end{bmatrix}$
Answer
D. $5 x_{1} + 4 x_{2} = 4 -5 x_{1} = -2 9 x_{1} - 9 x_{2} = 6 5 x_{1}$ 5x1 $+ + 4 x_{2}$ $= 4 4 -5 x_{1}$ -5x1 $= -2$ $9 x_{1}$ - $9 x_{2}$ $= 6 6$
1.3.6
Write a system of equations that is equivalent to the given vector equation. $x_{1} \begin{bmatrix} 5 \\ -4 \end{bmatrix} + x_{2} \begin{bmatrix} 9 \\ 5 \end{bmatrix} + x_{3} \begin{bmatrix} -4 \\ 1 \end{bmatrix}$=$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Answer
A. $5 x_{1}$ $+ + 9 x_{2}$ - $4 x_{3}$ $= 0 -4 x_{1}$ -4x1 $+ + 5 x_{2}$ $+ + x_{3}$ $= 0$
1.3.7
Use the accompanying figure to write each vector listed as a linear combination of u and v. Vectors a, w, x, and z
Answer
Write a as a linear combination of u and v. $a =$ (\square) $1 u +$ (-2) v (Type integers or decimals.) Part 2 Write w as a linear combination of u and v. $w =$ (\square) $-1 u +$ (2) v (Type integers or decimals.) Part 3 Write x as a linear combination of u and v. $x =$ (\square) $-2 u +$ (2) v (Type integers or decimals.) Part 4 Write z as a linear combination of u and v. $z =$ (\square) $-3 u +$ (4) v (Type integers or decimals.)
1.3.9
Write a vector equation that is equivalent to the given system of equations. $x_{2} + 4 x_{3} = 0 4 x_{1} + 7 x_{2} - x_{3} = 0$ negative $x_{1} + 3 x_{2} - 6 x_{3} = 0$
Answer
D. $x_{1} \begin{bmatrix} 0 \\ 4 \\ -1 \end{bmatrix}$ $+ +x_{2} \begin{bmatrix} 1 \\ 7 \\ 3 \end{bmatrix} 1 7 3 + +x_{3}$
1.3.11
Determine if b is a linear combination of $a_{1}, a_{2}$, and $a_{3}$. $a_{1} = \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix}$, $a_{2} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $a_{3} = \begin{bmatrix} 5 \\ -4 \\ 22 \end{bmatrix}$, $b = \begin{bmatrix} 4 \\ -3 \\ 18 \end{bmatrix}$
Answer
A. Vector b is a linear combination of $a_{1}, a_{2}$, and $a_{3}$. The pivots in the corresponding echelon matrix are in the first entry in the first column and the second entry in the second column.
1.3.13
Determine if b is a linear combination of the vectors formed from the columns of the matrix A. $A = \begin{bmatrix} 1 & -6 & 5 \\ 0 & 5 & 6 \\ -3 & 18 & -15 \end{bmatrix}$, $b = \begin{bmatrix} 4 \\ -6 \\ -4 \end{bmatrix}$
Answer
A. Choose the correct answer below. Your answer is correct
1.3.14
Determine if b is a linear combination of the vectors formed from the columns of the matrix A. $A = \begin{bmatrix} 1 & -6 & -4 \\ 0 & 4 & 3 \\ 3 & -18 & 11 \end{bmatrix}$,$\mathbf{b}$=$\begin{bmatrix} 12 \\ -5 \\ 7 \end{bmatrix}$
Answer
A. The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.
1.3.17
Let $a_{1} = \begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}$, $a_{2} = \begin{bmatrix} -7 \\ -31 \\ 3 \end{bmatrix}$, and $b = \begin{bmatrix} 5 \\ 13 \\ h \end{bmatrix}$. For what value(s) of h is b in the plane spanned by $a_{1}$ and $a_{2}$ ?
Answer
The value(s) of h is(are) 7. (Use a comma to separate answers as needed.)
1.3.23
Determine whether the statement below is true or false. Justify the answer. Another notation for the vector$\begin{bmatrix} -4 \\ 3 \end{bmatrix}$is$\begin{bmatrix} -4 & 3 \end{bmatrix}$
Answer
B. The statement is false. The alternative notation for a (column) vector is (- -4, 3), using parentheses and a comma.
1.3.27
Determine whether the statement below is true or false. Justify the answer. An example of a linear combination of vectors$\mathbf{v}_{1}$and$\mathbf{v}_{2}$is the vector one half$\mathbf{v}_{1}$
Answer
B. The statement is true because one half$\mathbf{v}_{1}$1 2 v1 $= $ one half$\mathbf{v}_{1} + 0 \mathbf{v}_{2}$1 2 v1+0v2.
1.3.29
Determine whether the statement below is true or false. Justify the answer. The solution set of the linear system whose augmented matrix is$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{b} \end{bmatrix}$is the same as the solution set of the equation $x_{1} \mathbf{a}_{1} + x_{2} \mathbf{a}_{2} + x_{3} \mathbf{a}_{3}$=$\mathbf{b}$
Answer
B. True. The augmented matrix for $x_{1} \mathbf{a}_{1} + x_{2} \mathbf{a}_{2} + x_{3} \mathbf{a}_{3}$=$\mathbf{b}$ is$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{b} \end{bmatrix}$a1 a2 a3 b.
1.3.31
Determine whether the statement below is true or false. Justify the answer. The set Span $\{ \mathbf{u}$,$\mathbf{v} \}$ is always visualized as a plane through the origin
Answer
D. The statement is false. It is often true, but Span $\{ \mathbf{u}$,$\mathbf{v} \}$ Span {u, v} is not a plane when$\mathbf{v}$ is a multiple of$\mathbf{u}$ or when$\mathbf{u}$ is the zero vector.
1.3.32
Determine whether the statement below is true or false. Justify the answer. Asking whether the linear system corresponding to an augmented matrix$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{b} \end{bmatrix}$has a solution amounts to asking whether$\mathbf{b}$is in Span $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$
Answer
A. The statement is true. The linear system corresponding to$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{b} \end{bmatrix}$a1 a2 a3 b has a solution when$\mathbf{b}$ can be written as a linear combination of$\mathbf{a}_{1}$,$\mathbf{a}_{2}$, and$\mathbf{a}_{3}$. This is equivalent to saying that$\mathbf{b}$ is in Span $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$ Sp
1.3.33
Let $A = \begin{bmatrix} 1 & 0 & -3 \\ 0 & 2 & -6 \\ -2 & 6 & 4 \end{bmatrix}$and$\mathbf{b}$=$\begin{bmatrix} 3 \\ 4 \\ -26 \end{bmatrix}$. Denote the columns of A by$\mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3}$, and let $W =$ Span $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$. a. Is$\mathbf{b}$in $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$ ? How many vectors are in $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$ ? b. Is$\mathbf{b}$in W? How many vectors are in W? c. Show that$\mathbf{a}_{2}$is in W. [ Hint: Row operations are unnecessary.]
Answer
D. No,$\mathbf{b}$ is not in $\{ \mathbf{a}_{1}$,$\mathbf{a}_{2}$,$\mathbf{a}_{3} \}$ since$\mathbf{b} b \ne \mathbf{a}_{1}$a1,$\mathbf{a}_{2}$, or$\mathbf{a}_{3}$.
1.3.37
Let$\mathbf{v}_{1}$,…., $\mathbf{v}_{k}$ be the points in $\mathbb{R}^{3}$ and suppose that for $j = 1$,…, k an object with mass $m_{j}$ is located at point $\mathbf{v}_{j}$. Physicists call such objects point masses. The total mass of the system of point masses is $m = m_{1} + \cdots + m_{k}$. The center of gravity (or center of mass) of the system is$\mathbf{v}$overbar $= \frac{1}{m} m_{1} \mathbf{v}_{1} + \cdots + m_{k} \mathbf{v}_{k}$. Point Mass$\mathbf{v}_{1} = (4, -4, 3) 8 g \mathbf{v}_{2} = (6, 2, -1) 20 g \mathbf{v}_{3}$= (-4, -3, -1) 8 g$\mathbf{v}_{4}$= (-9, 6, 6) 4 g Compute the center of gravity of the system consisting of the point masses above
Answer
The center of gravity is at$\mathbf{v}$overbar = (\square, \square, \square) $\frac{21}{10}$, one fifth, one half (Simplify your answers.)
1.3.42
Use the vector$\mathbf{u} = (u_{1}$,…, $u_{n})$ to verify the following algebraic properties of $\mathbb{R}^{n}$. a.$\mathbf{u}$+ (-$\mathbf{u}$) = (-$\mathbf{u}$) +$\mathbf{u}$=$\mathbf{0}$b. c(d$\mathbf{u}$) = (cd)$\mathbf{u}$for all scalars c and d See Sample Answer
Answer not captured
Open this one in MyLab review directly.
1.3
Find the indicated vector. Let $u = \begin{bmatrix} 6 \\ -8 \end{bmatrix}$, $v = \begin{bmatrix} 6 \\ 9 \end{bmatrix}$. Find $u - v$
Answer
D. $\begin{bmatrix} 0 \\ -17 \end{bmatrix}$
1.3.28
Determine whether the statement below is true or false. Justify the answer. The weights $c_{1}$,…, $c_{p}$ in a linear combination $c_{1} \mathbf{v}_{1} + \cdots + c_{p} \mathbf{v}_{p}$ cannot all be zero
Answer
C. The statement is false. Setting all the weights equal to zero results in the vector$\mathbf{0}$.
§1.4 The Matrix Equation Ax = b
1.4.1
Compute the product using (a) the definition where A x is the linear combination of the columns of A using the corresponding entries in x as weights, and (b) the row-vector rule for computing A x. If a product is undefined, explain why.$\begin{bmatrix} -1 & 2 \\ 2 & 9 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ -10 \\ 6 \end{bmatrix}$
Answer
B. The matrix-vector Ax is not defined because the number of columns in matrix A does not match the number of entries in the vector x.
1.4.4
Compute the product using the methods below. If a product is undefined, explain why. a. The definition where A x is the linear combination of the columns of A using the corresponding entries in x as weights. b. The row-vector rule for computing A x.$\begin{bmatrix} 7 & 5 & -6 \\ 3 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$
Answer
B. $x_{1} \mathbf{a}_{1} + x_{2} \mathbf{a}_{2} + \cdots + x_{n} \mathbf{a}_{n}$ =($x_{1}$)$\mathbf{a}_{1}$ +($x_{2}$)$\mathbf{a}_{2}$ +($x_{3}$)$\mathbf{a}_{3}$ $= $ (1)$\begin{bmatrix} 7 \\ 3$ (2) Start 2
1.4.6
Use the definition of A x to write the matrix equation as a vector equation.$\begin{bmatrix} 5 & -6 \\ -9 & 6 \\ -8 & 0 \\ -8 & -5 \end{bmatrix} \begin{bmatrix} -3 \\ 3 \end{bmatrix}$=$\begin{bmatrix} -33 \\ 45 \\ 24 \\ 9 \end{bmatrix}$
Answer
The matrix equation written as a vector equation is -3$\begin{bmatrix} 5 \\ -9 \\ -8 \\ -8 \end{bmatrix} + 3 \begin{bmatrix} -6 \\ 6 \\ 0 \\ -5 \end{bmatrix}$=$\begin{bmatrix} -33 \\ 45 \\ 24 \\ 9 \end{bmatrix}$
1.4.8
Use the definition of A x to write the vector equation as a matrix equation. $z_{1} \begin{bmatrix} 3 \\ 3 \end{bmatrix} + z_{2} \begin{bmatrix} -1 \\ -4 \end{bmatrix} + z_{3} \begin{bmatrix} -1 \\ 4 \end{bmatrix} + z_{4} \begin{bmatrix} -2 \\ 1 \end{bmatrix}$=$\begin{bmatrix} -2 \\ 2 \end{bmatrix}$
Answer
The vector equation written as a matrix equation is$\begin{bmatrix} 3 & -1 & -1 & -2 \\ 3 & -4 & 4 & 1 \end{bmatrix} \begin{bmatrix} z_{1} \\ z_{2} \\ z_{3} \\ z_{4} \end{bmatrix}$=$\begin{bmatrix} -2 \\ 2 \end{bmatrix}$
1.4.9
Write the system first as a vector equation and then as a matrix equation. $3 x_{1} + x_{2} - 3 x_{3} = 5 7 x_{2} + 4 x_{3} = 0$
Answer
A. $x_{1} \begin{bmatrix} 3 \\ 0 \end{bmatrix} + x_{2} \begin{bmatrix} 1 \\ 7 \end{bmatrix} + x_{3} \begin{bmatrix} -3 \\ 4 \end{bmatrix}$=$\begin{bmatrix} 5 \\ 0 \end{bmatrix} x_{1}$ x1$\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ p
1.4.10
Write the system first as a vector equation and then as a matrix equation. $3 x_{1} - x_{2} = 3 10 x_{1} + 5 x_{2} = 3 6 x_{1} - x_{2} = 1$
Answer
B. $x_{1} \begin{bmatrix} 3 \\ 10 \\ 6 \end{bmatrix} + x_{2} \begin{bmatrix} -1 \\ 5 \\ -1 \end{bmatrix}$=$\begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix} x_{1}$ x1
1.4.12
Given A and b to the right, write the augmented matrix for the linear system that corresponds to the matrix equation $A x = b$. Then solve the system and write the solution as a vector. $A = \begin{bmatrix} 1 & 5 & -7 \\ 1 & 4 & 2 \\ -4 & -7 & 2 \end{bmatrix}$, $b = \begin{bmatrix} -2 \\ 13 \\ -5 \end{bmatrix}$
Answer
Write the augmented matrix for the linear system that corresponds to the matrix equation $A x = b$.$\begin{bmatrix} 1 & 5 & -7 & -2 \\ 1 & 4 & 2 & 13 \\ -4 & -7 & 2 & -5 \end{bmatrix}$Part 2 Solve the system and write the solution as a vector. $x = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$=$\begin{bmatrix} -3 \\ 3 \\ 2 \end{bmatrix}$
1.4.13
Let $u = \begin{bmatrix} -21 \\ 27 \\ 6 \end{bmatrix}$and $A = \begin{bmatrix} 4 & -6 \\ -3 & 7 \\ 1 & 1 \end{bmatrix}$. Is u in the plane in $\mathbb{R}^{3}$ spanned by the columns of A? Why or why not?
Answer
A. Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer for each matrix element.) Your answer is correct
1.4.14
Let $u = \begin{bmatrix} 1 \\ -1 \\ 9 \end{bmatrix}$and $A = \begin{bmatrix} 5 & 8 & 12 \\ 1 & 2 & 2 \\ 2 & 1 & 7 \end{bmatrix}$. Is u in the subset of $\mathbb{R}^{3}$ spanned by the columns of A? Why or why not?
Answer
B. No, the reduced echelon form of the augmented matrix is$\begin{bmatrix} 1 & 0 & 4 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ which is an inconsistent system.
1.4.15
Let $A = \begin{bmatrix} 2 & -4 \\ -8 & 16 \end{bmatrix}$and $b = \begin{bmatrix} b_{1} \\ b_{2} \end{bmatrix}$. Show that the equation $A x = b$ does not have a solution for some choices of b, and describe the set of all b for which $A x = b$ does have a solution
Answer
A. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row.
1.4.16
Let $A = \begin{bmatrix} 1 & -3 & -2 \\ -4 & 4 & 0 \\ 3 & -1 & 2 \end{bmatrix}$and $b = \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \end{bmatrix}$. Show that the equation $A x = b$ does not have a solution for all possible b, and describe the set of all b for which $A x = b$ does have a solution
Answer
D. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row.
1.4.17
How many rows of A contain a pivot position? Does the equation $A x = b$ have a solution for each b in $\mathbb{R}^{4}$ ? $A = \begin{bmatrix} 2 & 4 & 1 & 2 \\ -2 & -2 & -2 & 2 \\ 0 & -4 & 2 & -8 \\ 2 & 0 & 3 & -3 \end{bmatrix}$
Answer
B. No, because A does not have a pivot position in every row.
1.4.21
Let$\mathbf{v}_{1}$=$\begin{bmatrix} 0 \\ 0 \\ -1 \\ 1 \end{bmatrix}$,$\mathbf{v}_{2}$=$\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}$, and$\mathbf{v}_{3}$=$\begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix}$. Does $\{ \mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3} \}$ span $\mathbb{R}^{4}$ ? Why or why not?
Answer
B. No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only three rows.
1.4.22
Let$\mathbf{v}_{1}$=$\begin{bmatrix} 0 \\ 0 \\ -3 \end{bmatrix}$,$\mathbf{v}_{2}$=$\begin{bmatrix} 0 \\ -2 \\ -6 \end{bmatrix}$, and$\mathbf{v}_{3}$=$\begin{bmatrix} 6 \\ -3 \\ 9 \end{bmatrix}$. Does $\{ \mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3} \}$ span $\mathbb{R}^{3}$ ? Why or why not?
Answer
D. Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.
1.4.23
Determine whether the statement below is true or false. Justify the answer. The equation $A x = b$ is referred to as a vector equation
Answer
A. This statement is false. The equation Ax =b is referred to as a matrix equation because A is a matrix.
1.4.24
Determine whether the statement below is true or false. Justify the answer. Every matrix equation $A x = b$ corresponds to a vector equation with the same solution set
Answer
A. Choose the correct answer below. Your answer is correct
1.4.26
Determine whether the statement below is true or false. Justify the answer. A vector b is a linear combination of the columns of a matrix A if and only if the equation $A x = b$ has at least one solution
Answer
D. This statement is true. The equation Ax =b has the same solution set as the equation $x_{1} \mathbf{a}_{1} + x_{2} \mathbf{a}_{2} + \cdots + x_{n} \mathbf{a}_{n} = \mathbf{b}$.
1.4.31
Determine whether the statement below is true or false. Justify the answer. If the columns of an $m \times n$ matrix A span $\mathbb{R}^{m}$, then the equation $A x = b$ is consistent for each b in $\mathbb{R}^{m}$
Answer
A. This statement is true. If the columns of A span $\mathbb{R}^{m}$, then the equation Ax =b has a solution for each b in $\mathbb{R}^{m}$.
1.4.32
Determine whether the statement below is true or false. Justify the answer. The solution set of a linear system whose augmented matrix is$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{b} \end{bmatrix}$is the same as the solution set of $A x = b$, if A=$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} \end{bmatrix}$
Answer
C. This statement is true. If A is an $m \times$ ×n matrix with columns$\begin{bmatrix} \mathbf{a}_{1} & \mathbf{a}_{2} & \cdots & \mathbf{a}_{n} \end{bmatrix}$a1 a2 ⋯ an, and b is a vector in $\mathbb{R}^{m}$, the matrix equation Ax =b has the same solution set as the system of linear equations who
1.4.42
Could a set of three vectors in $\mathbb{R}^{4}$ span all of $\mathbb{R}^{4}$ ? Explain. What
Answer
A. No. The matrix A whose columns are the three vectors has four rows. To have a pivot in each row, A would have to have at least four columns (one for each pivot).
§1.5 Solution Sets of Linear Systems
1.5.1
Determine if the system has a nontrivial solution. Try to use as few row operations as possible. $6 x_{1} - 3 x_{2} + 15 x_{3} = 0 -6 x_{1} - 9 x_{2} - 6 x_{3} = 0 12 x_{1} + 6 x_{2} + 21 x_{3} = 0$
Answer
A. The system has a nontrivial solution.
1.5.2
Determine if the system has a nontrivial solution. Try to use as few row operations as possible. $x_{1} - 4 x_{2} + 6 x_{3} = 0 - 2 x_{1} + 4 x_{2} - 3 x_{3} = 0 x_{1} +$ plus $10 x_{3} = 0$
Answer
B. The system has only a trivial solution.
1.5.3
Determine if the system has a nontrivial solution. Try to use as few row operations as possible. $-4 x_{1} + 6 x_{2} - 5 x_{3} = 0 -8 x_{1} + 9 x_{2} + 4 x_{3} = 0$
Answer
A. The system has a nontrivial solution.
1.5.8
Describe all solutions of $A x = 0$ in parametric vector form, where A is row equivalent to the given matrix.$\begin{bmatrix} 1 & -4 & -6 & 4 \\ 0 & 1 & 2 & -7 \end{bmatrix}$
Answer
$x = x_{3} \begin{bmatrix} -2 \\ -2 \\ 1 \\ 0 \end{bmatrix} + x_{4} \begin{bmatrix} 24 \\ 7 \\ 0 \\ 1 \end{bmatrix}$(Type an integer or fraction for each matrix element.)
1.5.9
Describe all solutions of $A x = 0$ in parametric vector form, where A is row equivalent to the given matrix.$\begin{bmatrix} 4 & -16 & 20 \\ -1 & 4 & -5 \end{bmatrix}$
Answer
$x = x_{2} \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} + x_{3} \begin{bmatrix} -5 \\ 0 \\ 1 \end{bmatrix}$(Type an integer or fraction for each matrix element.)
1.5.10
Describe all solutions of $A x = 0$ in parametric vector form, where A is row equivalent to the given matrix.$\begin{bmatrix} 1 & 4 & 0 & -2 \\ 4 & 16 & 0 & -8 \end{bmatrix}$
Answer
$x = x_{2} \begin{bmatrix} -4 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_{3} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_{4} \begin{bmatrix} 2 \\ 0 \\ 0 \\ 1 \end{bmatrix}$(Type an integer or fraction for each matrix element.)
1.5.17
Suppose the solution set of a certain system of linear equations can be described as $x_{1} = 4 + 5 x_{3}, x_{2} = -5 - 7 x_{3}$, with $x_{3}$ free. Use vectors to describe this set as a line in $\mathbb{R}^{3}$
Answer
Geometrically, the solution set is a line through$\begin{bmatrix} 4 \\ -5 \\ 0 \end{bmatrix}$parallel to$\begin{bmatrix} 5 \\ -7 \\ 1 \end{bmatrix}$
1.5.23
Find the parametric equation of the line through a parallel to b, using t as the parameter. $a = \begin{bmatrix} -5 \\ 2 \end{bmatrix}$, $b = \begin{bmatrix} -6 \\ 3 \end{bmatrix}$
Answer
$x = \begin{bmatrix} -5 \\ 2 \end{bmatrix} + t \begin{bmatrix} -6 \\ 3 \end{bmatrix}$(Type an integer or a simplified fraction for each matrix element.)
1.5.27
Determine whether the statement below is true or false. Justify the answer. A homogeneous equation is always consistent
Answer
B. The statement is true. A homogenous equation can be written in the form Ax =0, where A is an $m \times$ ×n matrix and 0 is the zero vector in $\mathbb{R}^{m}$. Such a system Ax =0 always has at least one solution, namely, $x =$ =0. Thus a homogenous equation is always consistent.
1.5.28
Determine whether the statement below is true or false. Justify the answer. If x is a nontrivial solution of $A x = 0$, then every entry in x is nonzero
Answer
A. The statement is false. A nontrivial solution of Ax =0 is a nonzero vector x that satisfies Ax =0. Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.
1.5.31
Determine whether the statement below is true or false. Justify the answer. The homogenous equation $A x = 0$ has the trivial solution if and only if the equation has at least one free variable
Answer
A. The statement is false. The homogeneous equation Ax =0 always has the trivial solution.
1.5.32
Determine whether the statement below is true or false. Justify the answer. The equation $A x = b$ is homogeneous if the zero vector is a solution
Answer
B. The statement is true. A system of linear equations is said to be homogeneous if it can be written in the form Ax =0, where A is an $m \times$ ×n matrix and 0 is the zero vector in $\mathbb{R}^{m}$. If the zero vector is a solution, then $b =$ =Ax =A0 =0.
1.5.34
Determine whether the statement below is true or false. Justify the answer. The effect of adding p to a vector is to move the vector in a direction parallel to p
Answer
B. The statement is true. Given v and p in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$, the effect of adding p to v is to move v in a direction parallel to the line through p and 0.
1.5.41
A is $a_{3} \times 3$ matrix with three pivot positions. (a) Does the equation $A x = 0$ have a nontrivial solution? (b) Does the equation $A x = b$ have at least one solution for every possible b ?
Answer
(a) Does the equation $A x = 0$ have a nontrivial solution? Your answer is correct. Part 2 (b) Does the equation $A x = b$ have at least one solution for every possible b ? Your answer is correct
1.5.42
Let A be $a_{3} \times 3$ matrix with two pivot positions. Use this information to answer parts (a) and (b) below
Answer
C. Yes. Since A has 2 pivots, there is one free variable. So A$\mathbf{x}$=$\mathbf{0}$Ax=0 has a nontrivial solution.
1.5.44
A is $a_{2} \times 5$ matrix with two pivot positions. (a) Does the equation $A x = 0$ have a nontrivial solution? (b) Does the equation $A x = b$ have at least one solution for every possible b ?
Answer
(a) Does the equation $A x = 0$ have a nontrivial solution? Your answer is correct. Part 2 (b) Does the equation $A x = b$ have at least one solution for every possible b ? Your answer is correct
§1.7 Linear Independence
1.7.2
Determine if the vectors are linearly independent. Justify your answer.$\begin{bmatrix} 7 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 9 \\ 3 \\ -12 \end{bmatrix}$,$\begin{bmatrix} 8 \\ 12 \\ -24 \end{bmatrix}$
Answer
The vector equation has only the trivial solution, so the vectors are linearly independent
1.7.4
Determine if the vectors are linearly independent.$\mathbf{v}_{1}$=$\begin{bmatrix} 1 \\ -2 \end{bmatrix}$,$\mathbf{v}_{2}$=$\begin{bmatrix} -2 \\ 4 \end{bmatrix}$
Answer
B. The vectors are not linearly independent because if $c_{1}$ $= 2 2$ and $c_{2}$ =1, both not zero, then $c_{1} \mathbf{v}_{1} + c_{2} \mathbf{v}_{2}$=$\mathbf{0}$.
1.7.5
Determine if the columns of the matrix form a linearly independent set. Justify your answer.$\begin{bmatrix} 0 & -8 & 16 \\ 3 & 1 & -14 \\ -1 & 5 & -5 \\ 1 & -5 & -2 \end{bmatrix}$
Answer
D. If A is the given matrix, then the augmented matrix$\begin{bmatrix} 0 & -8 & 16 & 0 \\ 3 & 1 & -14 & 0 \\ -1 & 5 & -5 & 0 \\ 1 & -5 & -2 & 0 \end{bmatrix}$
1.7.6
Determine if the columns of the matrix form a linearly independent set. Justify your answer.$\begin{bmatrix} -2 & -1 & 0 \\ 0 & -1 & 4 \\ 1 & 1 & -8 \\ 2 & 1 & -16 \end{bmatrix}$
Answer
D. If A is the given matrix, then the augmented matrix$\begin{bmatrix} -2 & -1 & 0 & 0 \\ 0 & -1 & 4 & 0 \\ 1 & 1 & -8 & 0 \\ 2 & 1 & -16 & 0 \end{bmatrix}$ -
1.7.7
Determine if the columns of the matrix form a linearly independent set.$\begin{bmatrix} 1 & 3 & -3 & 1 \\ 2 & 7 & -4 & -2 \\ 2 & 8 & 1 & -12 \end{bmatrix}$
Answer
A. The columns of the matrix do not form a linearly independent set because the set contains more vectors, 4, than there are entries in each vector, 3. (Type whole numbers.)
1.7.8
Determine if the columns of the matrix form a linearly independent set. Justify your answer.$\begin{bmatrix} 1 & -2 & 3 & 4 \\ -2 & 4 & -6 & 4 \\ 0 & 1 & -1 & 5 \end{bmatrix}$
Answer
C. The columns of the matrix do not form a linearly independent set because the set contains more vectors than there are entries in each vector.
1.7.15
Determine by inspection whether the vectors are linearly independent. Justify your answer.$\begin{bmatrix} 4 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 3 \\ 8 \end{bmatrix}$,$\begin{bmatrix} 1 \\ 3 \end{bmatrix}$,$\begin{bmatrix} -1 \\ 6 \end{bmatrix}$
Answer
D. The set is linearly dependent because there are four vectors but only two entries in each vector.
1.7.16
Determine by inspection whether the vectors are linearly independent. Justify your answer.$\begin{bmatrix} 3 \\ -9 \\ 6 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -6 \\ 4 \end{bmatrix}$
Answer
A. The set of vectors is linearly dependent because two thirds $2 3 \times$ the first vector is equal to the second vector. (Type an integer or a simplified fraction.)
1.7.17
Determine by inspection whether the vectors are linearly independent. Justify your answer.$\begin{bmatrix} 6 \\ 4 \\ -1 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -3 \\ 3 \\ 4 \end{bmatrix}$
Answer
C. The set of vectors is linearly dependent because one of the vectors is the zero vector.
1.7.19
Determine by inspection whether the vectors are linearly independent. Justify your answer.$\begin{bmatrix} -6 \\ -12 \\ -3 \end{bmatrix}$,$\begin{bmatrix} 2 \\ 4 \\ -1 \end{bmatrix}$
Answer
C. The set is linearly independent because neither vector is a multiple of the other vector. Two of the entries in the first vector are -3 $\times$ the corresponding entry in the second vector. But this multiple does not work for the third entries.
1.7.21
Determine whether the statement below is true or false. Justify the answer. The columns of a matrix A are linearly independent if the equation $A x = 0$ has the trivial solution
Answer
C. The statement is false. For every matrix A, Ax =0 has the trivial solution. The columns of A are independent only if the equation has no solution other than the trivial solution.
1.7.22
Determine whether the statement below is true or false. Justify the answer. Two vectors are linearly dependent if and only if they lie on a line through the origin
Answer
A. The statement is true. Two vectors are linearly dependent if one of the vectors is a multiple of the other. Two such vectors will lie on the same line through the origin.
1.7.23
Determine whether the statement below is true or false. Justify the answer. If S is a linearly dependent set, then each vector is a linear combination of the other vectors in S
Answer
D. The statement is false. If an indexed set of vectors, S, is linearly dependent, then it is only necessary that one of the vectors is a linear combination of the other vectors in the set.
1.7.24
Determine whether the statement below is true or false. Justify the answer. If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent
Answer
D. The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.
1.7.25
Determine whether the statement below is true or false. Justify the answer. The columns of any $4 \times 5$ matrix are linearly dependent
Answer
B. The statement is true. $A_{4} \times$ ×5 matrix has more columns than rows, and if a set contains more vectors than there are entries in each vector, then the set is linearly dependent.
1.7.26
Determine whether the statement below is true or false. Justify the answer. If x and y are linearly independent, and if z is in Span $\{ \mathbf{x}$,$\mathbf{y} \}$, then $\{ \mathbf{x}$,$\mathbf{y}$,$\mathbf{z} \}$ is linearly dependent
Answer
D. The statement is true. Since z is in Span $\{ \mathbf{x}$,$\mathbf{y} \}$ {x, y}, z is a linear combination of x and y. Since z is a linear combination of x and y, the set $\{ \mathbf{x}$,$\mathbf{y}$,$\mathbf{z} \}$ {x, y, z} is linearly dependent.
1.7.33
Suppose A is $a_{7} \times 5$ matrix. How many pivot columns must A have if its columns are linearly independent? Why?
Answer
B. The matrix must have 5 pivot columns. Otherwise, the equation A$\mathbf{x} x = \mathbf{0}$0 would have a free variable, in which case the columns of A would be linearly dependent.
1.7.34
Suppose A is $a_{5} \times 7$ matrix. How many pivot columns must A have if its columns span $\mathbb{R}^{5}$ ? Why?
Answer
A. The matrix must have 5 pivot columns. The statements “A has a pivot position in every row” and “the columns of A span $R \mathbb{R}^{5}$ 5” are logically equivalent.
1.7.37
Given $A = \begin{bmatrix} 4 & 1 & 3 \\ -3 & 3 & -6 \\ -8 & -1 & -7 \\ 5 & 0 & 5 \end{bmatrix}$, observe that the first column is the sum of the second and third columns. Find a nontrivial solution of A$\mathbf{x}$=$\mathbf{0}$without performing row operations. [Hint: Write A$\mathbf{x}$=$\mathbf{0}$as a vector equation.]
Answer
$x = \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$
1.7.38
Given $A = \begin{bmatrix} 4 & 14 & 5 \\ -7 & -15 & -4 \\ -2 & -10 & -4 \end{bmatrix}$, observe that the first column + twice the third column = the second column. Find a nontrivial solution of A$\mathbf{x}$=$\mathbf{0}$without performing row operations. [Hint: Write A$\mathbf{x}$=$\mathbf{0}$as a vector equation.]
Answer
$x = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$
§2.1 Matrix Operations
2.1.1
Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let $A = \begin{bmatrix} 2 & 0 & -1 \\ 4 & -5 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 8 & -4 & 1 \\ 2 & -5 & -4 \end{bmatrix}$, $C = \begin{bmatrix} 1 & 3 \\ -1 & 1 \end{bmatrix}$, and $D = \begin{bmatrix} 2 & 4 \\ -2 & 4 \end{bmatrix}$. -3 A, B -3 A, AC, CD
Answer
A. -3A $= \begin{bmatrix} -6 & 0 & 3 \\ -12 & 15 & -12 \end{bmatrix}$ (Simplify your answer.)
2.1.2
Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let $A = \begin{bmatrix} 2 & 0 & -3 \\ 3 & -4 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 8 & -5 & 1 \\ 1 & -4 & -3 \end{bmatrix}$, $C = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix}$, $D = \begin{bmatrix} 4 & 6 \\ -1 & 5 \end{bmatrix}$, and $E = \begin{bmatrix} -4 \\ 3 \end{bmatrix}$. $A + 3 B, 2 C - 3 E$, DB, EB
Answer
A. $A + + 3$ 3B $= \begin{bmatrix} 26 & -15 & 0 \\ 6 & -16 & -5 \end{bmatrix}$ (Simplify your answer.)
2.1.5
Compute the product AB by the definition of the product of matrices, where A$\mathbf{b}_{1}$and A$\mathbf{b}_{2}$are computed separately, and by the row-column rule for computing AB. $A = \begin{bmatrix} -2 & 4 \\ 1 & 5 \\ 6 & -3 \end{bmatrix}$, $B = \begin{bmatrix} 3 & -1 \\ -3 & 2 \end{bmatrix}$
Answer
D. $-2 (3) + 4$ (-3) $-2 (3) + 4$ (-3)
2.1.7
If a matrix A is $6 \times 9$ and the product AB is $6 \times 7$, what is the size of B?
Answer
The size of B is $9 \times 7$
2.1.8
How many rows does B have if BC is $a_{5} \times 2$ matrix?
Answer
Matrix B has 5 rows
2.1.9
Let $A = \begin{bmatrix} 3 & 2 \\ -1 & 2 \end{bmatrix}$and $B = \begin{bmatrix} 2 & 8 \\ -4 & k \end{bmatrix}$. What value(s) of k, if any, will make AB = BA?
Answer
A. $k = -2$ (Use a comma to separate answers as needed.)
2.1.10
Let $A = \begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix}$, $B = \begin{bmatrix} 11 & 1 \\ 8 & 7 \end{bmatrix}$, and $C = \begin{bmatrix} 2 & -8 \\ 2 & 1 \end{bmatrix}$. Verify that AB = AC and yet B not = C
Answer
B. Show the calculations that are used to find the entries for matrix AB. Choose the correct answer below. Your answer is correct. Part 2 Show the calculations that are used to find the entries for matrix AC. Choose the correct answer below. Your answer is correct. Part 3 Verify that AB = AC by simplifying. AB = AC =$\begin{bmatrix} -2 & -19 \\ 4 & 38 \end{bmatrix}$(Type an integer or decimal for each matrix element.)
2.1.15
Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer. If A and B are $2 \times 2$ with columns$\mathbf{a}_{1}$,$\mathbf{a}_{2}$, and$\mathbf{b}_{1}$,$\mathbf{b}_{2}$, respectively, then AB =$\begin{bmatrix} \mathbf{a}_{1} \mathbf{b}_{1} & \mathbf{a}_{2} \mathbf{b}_{2} \end{bmatrix}$
Answer
Fill in the blanks. The statement is false. The definition of matrix multiplication states that if A is an $m \times n$ matrix and B is an $n \times p$ matrix with columns$\mathbf{b}_{1}$,…, $\mathbf{b}_{p}$, then AB = A$\begin{bmatrix} \mathbf{b}_{1} & \mathbf{b}_{2} &... & \mathbf{b}_{p} \end{bmatrix}$=$\begin{bmatrix} A \mathbf{b}_{1} & A \mathbf{b}_{2} &... & A \mathbf{b}_{p} \end{bmatrix}$
2.1.16
Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer. If A and B are $3 \times 3$ matrices and $B = \begin{bmatrix} \mathbf{b}_{1} & \mathbf{b}_{2} & \mathbf{b}_{3} \end{bmatrix}$, then AB =$\begin{bmatrix} A \mathbf{b}_{1} + A \mathbf{b}_{2} + A \mathbf{b}_{3} \end{bmatrix}$
Answer
A. The statement is false. The matrix$\begin{bmatrix} A \mathbf{b}_{1} + A \mathbf{b}_{2} + A \mathbf{b}_{3} \end{bmatrix}$Ab1+Ab2+Ab3 is $a_{3} \times$ ×1 matrix, and AB must be $a_{3} \times$ ×3 matrix. The + signs should be spaces between the 3 columns.
2.1.19
Let A, B, and C be arbitrary matrices for which the indicated sums and products are defined. Determine whether the statement below is true or false. Justify the answer. AB + AC = A(B $+ C)$
Answer
B. The statement is true. The distributive law for matrices states that A(B $+ +C) =$ =AB + +AC.
2.1.20
Let A and B be arbitrary matrices for which the indicated sum is defined. Determine whether the statement below is true or false. Justify the answer. $A^{Upper} T + B^{Upper} T = (A + B)^{Upper}$ T
Answer
A. The statement is true. The transpose property states that $(A + B)^{Upper}$ T (A+B)T $= A^{Upper} T + B^{Upper}$ T AT+BT.
2.1.21
Let A, B, and C be arbitrary matrices for which the indicated products are defined. Determine whether the statement below is true or false. Justify the answer. (AB)C = (AC)B
Answer
C. Choose the correct answer below. Your answer is correct
2.1.22
Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer. $(AB)^{Upper} T = A^{Upper} T B^{Upper}$ T
Answer
B. The statement is false. The transpose of the product of two matrices is the product of the transposes of the individual matrices in reverse order, or $(AB)^{Upper}$ T (AB)T $= B^{Upper} T A^{Upper}$ T BTAT.
2.1.23
Determine whether the statement below is true or false. Justify the answer. The transpose of a product of matrices = the product of their transposes in the same order
Answer
C. The statement is false. The transpose of a product of matrices = the product of their transposes in the reverse order.
2.1.24
Determine whether the statement below is true or false. Justify the answer. The transpose of a sum of matrices = the sum of their transposes
Answer
C. The statement is true. This is a generalized statement that follows from the theorem $(A + B)^{Upper} T = A^{Upper} T + B^{Upper}$ T (A+B)T=AT+BT.
2.1.26
Suppose the first two columns,$\mathbf{b}_{1}$and$\mathbf{b}_{2}$, of B are equal. What can you say
Answer
B. The first two columns of AB are A$\mathbf{b}_{1}$ and A$\mathbf{b}_{2}$. They are equal since$\mathbf{b}_{1}$ and$\mathbf{b}_{2}$ are equal.
2.1.27
Suppose the fourth column of B is the sum of the first and last columns. What can be said
Answer
C. The fourth column of AB is the sum of the first and last columns of AB. If B is$\begin{bmatrix} \mathbf{b}_{1} & \mathbf{b}_{2} &... & \mathbf{b}_{p} \end{bmatrix}$b1 b2… bp, then the fourth column of AB is A$\mathbf{b}_{4}$Ab4 by definition. It is given that$\mathbf{b}_{4}$=$\mathbf{b}_{1} + \mathbf{b}_{p}$. By matrix-
2.1.47
Let $S = \begin{bmatrix} 0 & 9 & 0 & 0 & 0 \\ 0 & 0 & 9 & 0 & 0 \\ 0 & 0 & 0 & 9 & 0 \\ 0 & 0 & 0 & 0 & 9 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$. Compute $S^{k}$ for $k = 2$,…, 6
Answer
S squared =$\begin{bmatrix} 0 & 0 & 81 & 0 & 0 \\ 0 & 0 & 0 & 81 & 0 \\ 0 & 0 & 0 & 0 & 81 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$Part 2 S cubed =$\begin{bmatrix} 0 & 0 & 0 & 729 & 0 \\ 0 & 0 & 0 & 0 & 729 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$Part $3 S^{4} = \begin{bmatrix} 0 & 0 & 0 & 0 & 6561 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$Part $4 S^{5} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$Part $5 S^{6} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
2.1.52
Use the matrix $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$to scrub the data in matrix N which contains dates of accidents at a certain airport. $N = \begin{bmatrix} 10 & 10 & 10 & 10 & 11 & 11 & 11 \\ 1 & 14 & 20 & 21 & 4 & 9 & 24 \\ 2019 & 2019 & 2019 & 2019 & 2019 & 2019 & 2019 \end{bmatrix}$
Answer
The data in matrix N has been scrubbed in matrix BN =$\begin{bmatrix} 10 & 10 & 10 & 10 & 11 & 11 & 11 \\ 1 & 14 & 20 & 21 & 4 & 9 & 24 \end{bmatrix}$. (Simplify your answer.)
2.1.4
Compute $A - 2 I_{3}$ and ($2 I_{3}$)A, where $A = \begin{bmatrix} 6 & -2 & 4 \\ -5 & 2 & -5 \\ -3 & 1 & 2 \end{bmatrix}$
Answer
$A - 2 I_{3} = \begin{bmatrix} 4 & -2 & 4 \\ -5 & 0 & -5 \\ -3 & 1 & 0 \end{bmatrix}$Part 2 ($2 I_{3}$)A =$\begin{bmatrix} 12 & -4 & 8 \\ -10 & 4 & -10 \\ -6 & 2 & 4 \end{bmatrix}$
§2.2 The Inverse of a Matrix
2.2.1
Find the inverse of the matrix.$\begin{bmatrix} 4 & 6 \\ 8 & 7 \end{bmatrix}$
Answer
A. The inverse matrix is$\begin{bmatrix} -\frac{7}{20} & three tenths \\ two fifths & -one fifth \end{bmatrix}$- 7 20 3 10 2 5 - 1 5. (Type an integer or simplified fraction for each matrix element.)
2.2.3
Find the inverse of the matrix.$\begin{bmatrix} 7 & 7 \\ -6 & -7 \end{bmatrix}$
Answer
A. The inverse matrix is$\begin{bmatrix} 1 & 1 \\ -six sevenths & -1 \end{bmatrix}$ - 6 7 -1. (Type an integer or simplified fraction for each matrix element.)
2.2.4
Find the inverse of the matrix.$\begin{bmatrix} -5 & -4 \\ 3 & 2 \end{bmatrix}$
Answer
A. The inverse matrix is$\begin{bmatrix} 1 & 2 \\ -three halves & -five halves \end{bmatrix}$ - 3 2 - 5 2. (Type an integer or simplified fraction for each matrix element.)
2.2.9
Let $A = \begin{bmatrix} 1 & 2 \\ 8 & 18 \end{bmatrix}$,$\mathbf{b}_{1}$=$\begin{bmatrix} 0 \\ 8 \end{bmatrix}$,$\mathbf{b}_{2}$=$\begin{bmatrix} 0 \\ -10 \end{bmatrix}$,$\mathbf{b}_{3}$=$\begin{bmatrix} 2 \\ 14 \end{bmatrix}$, and$\mathbf{b}_{4}$=$\begin{bmatrix} 4 \\ 22 \end{bmatrix}$. a. Find $A^{-1}$ and use it solve the four equations $A x = \mathbf{b}_{1}$, $A x = \mathbf{b}_{2}$, $A x = \mathbf{b}_{3}$, and $A x = \mathbf{b}_{4}$. b. The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A$\mathbf{b}_{1} \mathbf{b}_{2} \mathbf{b}_{3} \mathbf{b}_{4}$]
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} 9 & -1 \\ -4 & one half \end{bmatrix}$. (Type an integer or simplified fraction for each matrix element.)
2.2.11
Determine whether the statement below is true or false. Justify the answer. In order for a matrix B to be the inverse of A, both equations AB = I and BA = I must be true
Answer
C. The statement is true. The product of a matrix and its inverse is the identity matrix.
2.2.12
Determine whether the statement below is true or false. Justify the answer. A product of invertible $n \times n$ matrices is invertible, and the inverse of the product is the product of their inverses in the same order
Answer
C. The statement is false. If A and B are invertible matrices, then $(AB)^{-1}$ $= B^{-1}$ $A^{-1}$.
2.2.13
Determine whether the statement below is true or false. Justify the answer. If A and B are $n \times n$ and invertible, then $A^{-1} B^{-1}$ is the inverse of AB
Answer
B. The statement is false. The inverse of AB is $B^{-1} A^{-1}$.
2.2.14
Determine whether the statement below is true or false. Justify the answer. If A is invertible, then the inverse of $A^{-1}$ is A itself
Answer
D. The statement is true. Since $A^{-1}$ is the inverse of $A, A^{-1}$ A-1A =I =A $A^{-1}$. Since $A^{-1}$ A =I =A $A^{-1}$, A is the inverse of $A^{-1}$.
2.2.15
Determine whether the statement below is true or false. Justify the answer. If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$and ab - cd not $= 0$, then A is invertible
Answer
D. The statement is false. If ad - -bc not = ≠0, then A is invertible.
2.2.16
Determine whether the statement below is true or false. Justify the answer. If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$and ad = bc, then A is not invertible
Answer
C. The statement is true. If ad =bc then ad - -bc =0, and $\frac{1}{ad -$ bc}$\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ ad-bc d -b -c a is undefined.
2.2.17
Determine whether the statement below is true or false. Justify the answer. If A is an invertible $n \times n$ matrix, then the equation A$\mathbf{x}$=$\mathbf{b}$is consistent for each$\mathbf{b}$in $\mathbb{R}^{n}$
Answer
D. The statement is true. Since A is invertible, $A^{-1}$ $\mathbf{b}$ exists for all$\mathbf{b}$ in $\mathbb{R}^{n}$. Define$\mathbf{x} x = A^{-1}$ A-1$\mathbf{b}$. Then A$\mathbf{x} x = \mathbf{b}$b.
2.2.19
Determine whether the statement below is true or false. Justify the answer. Each elementary matrix is invertible
Answer
D. The statement is true. Since each elementary matrix corresponds to a row operation, and every row operation is reversible, every elementary matrix has an inverse matrix.
2.2.25
Suppose A, B, and C are invertible $n \times n$ matrices. Show that ABC is also invertible by introducing a matrix D such that (ABC)D = I and D(ABC) = I
Answer
C. $A^{-1}$, $B^{-1}$, and $C^{-1}$ exist.
2.2.31
Explain why the columns of an $n \times n$ matrix A are linearly independent when A is invertible
Answer
B. If A is invertible, then the equation A$\mathbf{x} x = \mathbf{0}$0 has the unique solution$\mathbf{x} x = \mathbf{0}$0. Since A$\mathbf{x} x = \mathbf{0}$0 has only the trivial solution, the columns of A must be linearly independent.
2.2.32
Explain why the columns of an $n \times n$ matrix A span $\mathbb{R}^{n}$ when A is invertible
Answer
C. Since A is invertible, for each$\mathbf{b}$ in $\mathbb{R}^{n}$ the equation A$\mathbf{x} x = \mathbf{b}$b has a unique solution. Since the equation A$\mathbf{x} x = \mathbf{b}$b has a solution for all$\mathbf{b}$ in $\mathbb{R}^{n}$, the columns of A span $\mathbb{R}^{n}$.
2.2.39
Find the inverse of the matrix, if it exists. Use the algorithm for finding $A^{-1}$ by row reducing A I ]. $A = \begin{bmatrix} 1 & 6 \\ 5 & 29 \end{bmatrix}$
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} -29 & 6 \\ 5 & -1 \end{bmatrix}$. (Type an integer or simplified fraction for each matrix element.)
2.2.40
Find the inverse of the matrix, if it exists. Use the algorithm for finding $A^{-1}$ by row reducing A I. $A = \begin{bmatrix} 4 & 3 \\ 2 & 7 \end{bmatrix}$
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} \frac{7}{22} & -\frac{3}{22} \\ -one eleventh & two elevenths \end{bmatrix}$ - 3 22 - 1 11 2 11. (Type an integer or simplified fraction for each matrix element.)
2.2.41
Find the inverse of the given matrix, if it exists. Use the algorithm for finding $A^{-1}$ by row reducing A I. $A = \begin{bmatrix} 1 & 0 & -2 \\ -3 & 1 & 4 \\ 3 & 2 & 4 \end{bmatrix}$
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} -two sevenths & -two sevenths & one seventh \\ \frac{12}{7} & five sevenths & one seventh \\ -nine fourteenths & -one seventh & one fourteenth \end{bmatrix}$- 2
2.2.42
Find the inverse of the given matrix, if it exists. $A = \begin{bmatrix} 1 & -2 & 1 \\ 3 & -5 & 2 \\ -2 & 5 & -3 \end{bmatrix}$
Answer
B. The matrix A does not have an inverse.
§2.3 Characterizations of Invertible Matrices
2.3.1
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 4 & 3 \\ -3 & -6 \end{bmatrix}$
Answer
D. The matrix is invertible because its determinant is not zero.
2.3.3
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 4 & 0 & 0 \\ -3 & -5 & 0 \\ 7 & 6 & -3 \end{bmatrix}$
Answer
B. The matrix is invertible. The given matrix has three pivot positions.
2.3.4
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 4 & 0 & -4 \\ 2 & 0 & 5 \\ -4 & 0 & 6 \end{bmatrix}$
Answer
D. The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.
2.3.5
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 0 & 3 & -4 \\ 2 & 0 & 2 \\ -8 & -9 & 4 \end{bmatrix}$
Answer
D. The matrix is not invertible. If the given matrix is A, A is not row equivalent to the $n \times n$ identity matrix.
2.3.6
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 1 & -4 & -6 \\ 0 & 4 & 2 \\ -4 & 13 & 0 \end{bmatrix}$
Answer
D. The matrix is invertible. The given matrix has 3 pivot positions.
2.3.8
Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.$\begin{bmatrix} 3 & 5 & 7 & 5 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
Answer
C. The matrix is invertible. The given matrix has 4 pivot positions.
2.3.11
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If the equation $A x = 0$ has only the trivial solution, then A is row equivalent to the $n \times n$ identity matrix
Answer
B. The statement is true. By the Invertible Matrix Theorem, if the equation Ax =0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the $n \times n$ identity matrix.
2.3.13
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If the columns of A span $\mathbb{R}^{n}$, then the columns are linearly independent
Answer
D. The statement is true. The Invertible Matrix Theorem states that if the columns of A span $\mathbb{R}^{n}$, then matrix A is invertible. Therefore, the columns are linearly independent.
2.3.14
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If the columns of A are linearly independent, then the columns of A span $\mathbb{R}^{n}$
Answer
C. The statement is true. By the Invertible Matrix Theorem, if the columns of A are linearly independent, then the columns of A must span $\mathbb{R}^{n}$.
2.3.15
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If A is an $n \times n$ matrix, then the equation A$\mathbf{x}$=$\mathbf{b}$has at least one solution for each$\mathbf{b}$in $\mathbb{R}^{n}$
Answer
D. The statement is false. By the Invertible Matrix Theorem, A$\mathbf{x} x = \mathbf{b}$b has at least one solution for each$\mathbf{b}$ in $\mathbb{R}^{n}$ only if a matrix is invertible.
2.3.16
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If the equation $A x = b$ has at least one solution for each b in $\mathbb{R}^{n}$, then the solution is unique for each b
Answer
D. The statement is true. By the Invertible Matrix Theorem, if Ax =b has at least one solution for each b in $\mathbb{R}^{n}$, then matrix A is invertible. If A is invertible, then according to the invertible matrix theorem the solution is unique for each b.
2.3.17
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If the equation $A x = 0$ has a nontrivial solution, then A has fewer than n pivot positions
Answer
A. The statement is true. By the Invertible Matrix Theorem, if the equation Ax =0 has a nontrivial solution, then matrix A is not invertible. Therefore, A has fewer than n pivot positions.
2.3.20
For this exercise assume that the matrices are all $n \times$ n. The statement in this exercise is an implication of the form “If “statement 1”, then “statement 2”.” Mark an implication as True if the truth of “statement 2” always follows whenever “statement 1” happens to be true. Mark the implication as False if “statement 2” is false but “statement 1” is true. Justify your answer. If there is a b in $\mathbb{R}^{n}$ such that the equation $A x = b$ is inconsistent, then the transformation$\mathbf{x}$maps to A$\mathbf{x}$is not one-to-one
Answer
A. The statement is true. According to the Invertible Matrix Theorem, if there is a b in $\mathbb{R}^{n}$ such that the equation Ax =b is inconsistent, then equation Ax =b does not have at least one solution for each b in $\mathbb{R}^{n}$ and this makes A not invertible.
2.3.21
An $m \times n$ upper triangular matrix is one whose entries below the main diagonal are zeros, as is shown in the matrix to the right. When is a square upper triangular matrix invertible? Justify your answer.$\begin{bmatrix} 3 & 4 & 7 & 4 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
Answer
D. A square upper triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the $n \times$ ×n matrix has n pivot positions.
2.3.30
Suppose H is an $n \times n$ matrix. If the equation $H x = c$ is inconsistent for some c in $\mathbb{R}^{n}$, what can you say
Answer
C. The statement that Hx =c is inconsistent for some c is equivalent to the statement that Hx =c has no solution for some c. From this, all of the statements in the Invertible Matrix Theorem are false, including the statement that Hx =0 has only the trivial solution. Thus, Hx =0 has a nontrivial solution.
§3.1 Introduction to Determinants
3.1.1
Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.$\begin{bmatrix} 3 & 0 & 4 \\ 3 & 3 & 3 \\ 0 & 5 & -1 \end{bmatrix}$
Answer
B. Using this expansion, the determinant is (3)(-18) -(0)(-3) +(4)(15) $= 6$.
3.1.3
Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. $\left| \begin{bmatrix} 3 & -3 & 8 \\ 8 & 5 & 3 \\ 5 & 8 & -5 \end{bmatrix} \right|$
Answer
B. Using this expansion, the determinant is (3)(-49) -(-3)(-55) +(8)(39).
3.1.4
Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. $\left| \begin{bmatrix} 2 & -2 & 4 \\ 3 & 1 & 3 \\ 1 & 5 & -1 \end{bmatrix} \right|$
Answer
C. Using this expansion, the determinant is (2)(-16) -(-2)(-6) +(4)(14) $= 12$.
3.1.5
Compute the determinant of the following matrix using a cofactor expansion across the first row. $\left| \begin{bmatrix} 3 & 7 & -2 \\ 2 & 0 & 6 \\ 4 & 5 & 3 \end{bmatrix} \right|$
Answer
D. Using this expansion, the determinant is (3)(-30) -(7)(-18) +(-2)($10 10) = 16$.
3.1.9
Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation. $\left| \begin{bmatrix} 3 & 0 & 0 & 4 \\ 2 & 7 & 2 & -2 \\ 3 & 0 & 0 & 0 \\ 1 & 2 & 1 & 7 \end{bmatrix} \right|$
Answer
$\left| \begin{bmatrix} 3 & 0 & 0 & 4 \\ 2 & 7 & 2 & -2 \\ 3 & 0 & 0 & 0 \\ 1 & 2 & 1 & 7 \end{bmatrix} \right| = 36$ (Simplify your answer.)
3.1.10
Compute the following determinant by cofactor expansions. At each step, choose the row or column that involves the least amount of computation. $\left| \begin{bmatrix} 2 & -6 & 3 & 4 \\ 0 & 0 & 4 & 0 \\ 1 & -2 & -3 & 3 \\ 5 & 0 & 5 & 6 \end{bmatrix} \right|$
Answer
The determinant is 152
3.1.11
Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation. $\left| \begin{bmatrix} 2 & 1 & -7 & 4 \\ 0 & -4 & 2 & -1 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 5 \end{bmatrix} \right|$
Answer
$\left| \begin{bmatrix} 2 & 1 & -7 & 4 \\ 0 & -4 & 2 & -1 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 5 \end{bmatrix} \right| = -40$ (Simplify your answer.)
3.1.13
Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation.$\begin{bmatrix} 3 & 0 & -8 & 3 & -6 \\ 0 & 0 & 4 & 0 & 0 \\ 8 & 3 & -6 & 4 & -8 \\ 5 & 0 & 4 & 4 & -4 \\ 0 & 0 & 7 & -1 & 5 \end{bmatrix}$
Answer
$\begin{bmatrix} 3 & 0 & -8 & 3 & -6 \\ 0 & 0 & 4 & 0 & 0 \\ 8 & 3 & -6 & 4 & -8 \\ 5 & 0 & 4 & 4 & -4 \\ 0 & 0 & 7 & -1 & 5 \end{bmatrix} = -36$ (Simplify your answer.)
3.1.14
Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation. $\left| \begin{bmatrix} 2 & 3 & 3 & 4 & 0 \\ 9 & 0 & -3 & 1 & 0 \\ 3 & -6 & 7 & 7 & 1 \\ 3 & 0 & 0 & 0 & 0 \\ 2 & 2 & 4 & 2 & 0 \end{bmatrix} \right|$
Answer
$\left| \begin{bmatrix} 2 & 3 & 3 & 4 & 0 \\ 9 & 0 & -3 & 1 & 0 \\ 3 & -6 & 7 & 7 & 1 \\ 3 & 0 & 0 & 0 & 0 \\ 2 & 2 & 4 & 2 & 0 \end{bmatrix} \right| = 0$ (Simplify your answer.)
3.1.17
The expansion of $a_{3} \times 3$ determinant can be remembered by this device. Write a second copy of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals. Add the downward diagonal products and subtract the upward products. Use this method to compute the following determinant.$\begin{bmatrix} 1 & 2 & -5 \\ -5 & 4 & 2 \\ -2 & -4 & -3 \end{bmatrix}$- minus $- +$ plus $+ a_{11} a_{12} a_{13} a_{11} a_{12} a_{21} a_{22} a_{23} a_{21} a_{22} a_{31} a_{32} a_{33} a_{31} a_{32}$
Answer
$\begin{bmatrix} 1 & 2 & -5 \\ -5 & 4 & 2 \\ -2 & -4 & -3 \end{bmatrix} = -182$
3.1.21
Explore the effects of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.$\begin{bmatrix} 4 & 8 \\ 9 & 2 \end{bmatrix}$,$\begin{bmatrix} 4 & 8 \\ 9 + 4 k & 2 + 8 k \end{bmatrix}$
Answer
C. Replace row 2 with $k \times$ row $1 +$ row 2.
3.1.25
Compute the determinant of the following elementary matrix.$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & d \\ 0 & 0 & 1 \end{bmatrix}$
Answer
$\left| \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & d \\ 0 & 0 & 1 \end{bmatrix} \right| = 1$ (Simplify your answer.)
3.1.26
Compute the determinant of the following elementary matrix.$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Answer
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = -1$ (Simplify your answer.)
3.1.29
Compute the determinant of the following elementary matrix.$\begin{bmatrix} 1 & 0 & 0 \\ 0 & z & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Answer
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & z & 0 \\ 0 & 0 & 1 \end{bmatrix} = z$ (Simplify your answer.)
3.1.37
Let $A = \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}$. Write 3 A. Is det (3 A) equal to 3 det (A) ?
Answer
A. No, det (3 A) det(3A) $\ne 3$ det (A) 3det(A). The value of det (3 A) det(3A) is -18, whereas the value of 3 det (A) 3det(A) is -6.
§3.2 Properties of Determinants
3.2.1
State which property of determinants is illustrated in this equation. $\left| \begin{bmatrix} -7 & 8 & -1 \\ 21 & -3 & 6 \\ -5 & 7 & -6 \end{bmatrix} \right| =$ negative $\left| \begin{bmatrix} 21 & -3 & 6 \\ -7 & 8 & -1 \\ -5 & 7 & -6 \end{bmatrix} \right|$
Answer
A. Choose the correct answer below. Your answer is correct
3.2.2
The equation below illustrates a property of determinants. State the property.$\begin{bmatrix} 3 & -6 & 9 \\ 3 & 5 & -5 \\ 1 & 3 & 3 \end{bmatrix} = 3 \begin{bmatrix} 1 & -2 & 3 \\ 3 & 5 & -5 \\ 1 & 3 & 3 \end{bmatrix}$
Answer
B. Multiplying a row by 3 multiplies the determinant by 3.
3.2.5
Find the determinant by row reduction to echelon form. $\left| \begin{bmatrix} 1 & 5 & -6 \\ -1 & -4 & -5 \\ 2 & 8 & 7 \end{bmatrix} \right|$
Answer
Use row operations to reduce the matrix to echelon form.$\begin{bmatrix} 1 & 5 & -6 \\ -1 & -4 & -5 \\ 2 & 8 & 7 \end{bmatrix}$~$\begin{bmatrix} 1 & 5 & -6 \\ 0 & 1 & -11 \\ 0 & 0 & -3 \end{bmatrix}$Part 2 Find the determinant of the given matrix. $\left| \begin{bmatrix} 1 & 5 & -6 \\ -1 & -4 & -5 \\ 2 & 8 & 7 \end{bmatrix} \right| = -3$ (Simplify your answer.)
3.2.6
Find the determinant by row reduction to echelon form. $\left| \begin{bmatrix} 5 & 5 & -5 \\ 6 & 7 & -7 \\ 2 & -3 & -5 \end{bmatrix} \right|$
Answer
Use row operations to reduce the matrix to echelon form.$\begin{bmatrix} 5 & 5 & -5 \\ 6 & 7 & -7 \\ 2 & -3 & -5 \end{bmatrix}$~$\begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & -8 \end{bmatrix}$Part 2 Find the determinant of the given matrix. $\left| \begin{bmatrix} 5 & 5 & -5 \\ 6 & 7 & -7 \\ 2 & -3 & -5 \end{bmatrix} \right| = -40$ (Simplify your answer.)
3.2.12
Combine the methods of row reduction and cofactor expansion to compute the determinant. $\left| \begin{bmatrix} -1 & 4 & 8 & 0 \\ 4 & 3 & 2 & 0 \\ 4 & 4 & 6 & 6 \\ 4 & 2 & 4 & 3 \end{bmatrix} \right|$
Answer
The determinant is 306. (Simplify your answer.)
3.2.14
Combine the methods of row reduction and cofactor expansion to compute the determinant. $\left| \begin{bmatrix} -1 & -5 & -4 & -1 \\ 0 & 4 & 8 & 0 \\ -3 & -5 & -4 & -1 \\ 6 & -4 & -4 & 0 \end{bmatrix} \right|$
Answer
The determinant is 32. (Simplify your answer.)
3.2.21
Use determinants to find out if the matrix is invertible.$\begin{bmatrix} -5 & 0 & 1 \\ 1 & -3 & -2 \\ 0 & -5 & -3 \end{bmatrix}$
Answer
The determinant of the matrix is 0. (Simplify your answer.) Part 2 Is the matrix invertible? Choose the correct answer below. Your answer is correct
3.2.22
Use determinants to find out if the matrix is invertible.$\begin{bmatrix} -25 & -2 & 3 \\ -5 & 6 & 6 \\ 0 & -10 & -9 \end{bmatrix}$
Answer
D. The matrix is invertible because the determinant of the matrix is not zero.
3.2.23
Use determinants to find out if the matrix is invertible.$\begin{bmatrix} 1 & -1 & -3 & 0 \\ 0 & 1 & 5 & 4 \\ 3 & -1 & -3 & 4 \\ -1 & 2 & 8 & 5 \end{bmatrix}$
Answer
B. The matrix is invertible.
3.2.29
Let A be an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If the columns of A are linearly dependent, then det $A = 0$
Answer
D. The statement is true. If the columns of A are linearly dependent, then A is not invertible.
3.2.33
Let A and B be $n \times n$ matrices. Determine whether the statement below is true or false. Justify the answer. det(A $+ B) =$ det $A +$ det B
Answer
D. The statement is false. If $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ then det(A $+ +B) =$ =0 and det $A +$ +det $B =$ =2.
3.2.34
Let A be an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. det $A^{-1} =$ (- 1)det A
Answer
A. The statement is false. det $A^{-1}$ =(det $A)^{-1}$ -1
3.2.42
Find a formula for det(rA) when A is an $n \times n$ matrix
Answer
A. det (rA) $= \mathbb{R}^{n} \times$ det A det (rA) $= \mathbb{R}^{n} \times$ det A det(rA) = rn•det A
3.2.43
Verify that det AB = (det A)(det B), where the matrices A and B are given below. $A = \begin{bmatrix} 2 & 0 \\ 4 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 5 & 0 \\ 5 & 10 \end{bmatrix}$
Answer
Calculate det A and det B. det $A = 2$, det $B = 50$ (Simplify your answers.) Part 2 Now calculate the product (det A)(det B). (det A)(det $B) = 100$ (Simplify your answer.) Part 3 Calculate the product of matrices AB. AB =$\begin{bmatrix} 10 & 0 \\ 25 & 10 \end{bmatrix}$(Type an integer or decimal for each matrix element.) Part 4 Now calculate the determinant of the product of matrices A and B. det (AB) $= 100$ (Simplify your answer.)
3.2.45
Let A and B be $3 \times 3$ matrices, with det $A = 2$ and det $B = -9$. Use properties of determinants to complete parts (a) through (e) below
Answer
a. Compute det AB. det AB $= -18$ (Type an integer or a fraction.) Part 2 b. Compute det 5A. det 5A $= 250$ (Type an integer or a fraction.) Part 3 c. Compute det $B^{Upper}$ T. det $B^{Upper} T = -9$ (Type an integer or a fraction.) Part 4 d. Compute det $A^{-1}$. det $A^{-1} =$ one half (Type an integer or a simplified fraction.) Part 5 e. Compute det A cubed. det A cubed $= 8$ (Type an integer or a fraction.)
§3.3 Cramer’s Rule; Volume
3.3.1
Use Cramer’s rule to compute the solutions of the system. $4 x_{1} + 4 x_{2} = 16 -3 x_{1} + 7 x_{2} = -42$
Answer
What is the solution of the system? $x_{1} = 7 x_{2} = -3$
3.3.2
Use Cramer’s rule to compute the solution of the system. $5 x_{1} + 3 x_{2} = 13 4 x_{1} + 6 x_{2} = 14$
Answer
What is the solution of the system? $x_{1} = 2 x_{2} = 1$ (Type integers or simplified fractions.)
3.3.3
Use Cramer’s rule to compute the solutions of the system. $9 x_{1} + 7 x_{2} = 13 7 x_{1} + 7 x_{2} = 9$
Answer
What is the solution of the system? $x_{1} = 2 x_{2} =$ negative five sevenths (Type integers or simplified fractions.)
3.3.5
Use Cramer’s rule to compute the solution of the system.$\begin{bmatrix} x_{1} & + & x_{2} & \, & \, & = & 3 \\ -2 x_{1} & \, & \, & + & 3 x_{3} & = & 0 \\ \, & \, & x_{2} & - & 3 x_{3} & = & 2 \end{bmatrix}$
Answer
$x_{1} =$ one third; $x_{2} =$ eight thirds; $x_{3} =$ two ninths (Type integers or simplified fractions.)
3.3.6
Use Cramer’s rule to compute the solution of the system.$\begin{bmatrix} 11 x_{1} & + & 3 x_{2} & + & x_{3} & = & 4 \\ 2 x_{1} & \, & \, & + & 5 x_{3} & = & 4 \\ -4 x_{1} & + & 4 x_{2} & \, & \, & = & 2 \end{bmatrix}$
Answer
$x_{1} =$ one eighth; $x_{2} =$ five eighths; $x_{3} =$ three fourths (Type integers or simplified fractions.)
3.3.12
Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix. $A = \begin{bmatrix} 1 & 1 & 4 \\ -3 & 3 & 2 \\ 0 & 2 & 4 \end{bmatrix}$
Answer
The adjugate of the given matrix is adj $A = \begin{bmatrix} 8 & 4 & -10 \\ 12 & 4 & -14 \\ -6 & -2 & 6 \end{bmatrix}$. (Type an integer or simplified fraction for each matrix element.) Part 2 The inverse of the given matrix is $A^{-1} = \begin{bmatrix} -2 & -1 & five halves \\ -3 & -1 & seven halves \\ three halves & one half & -three halves \end{bmatrix}$. (Type an integer or simplified fraction for each matrix element.)
3.3.13
Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix. $A = \begin{bmatrix} 4 & 6 & 3 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{bmatrix}$
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} -one fifth & -three fifths & six fifths \\ one fifth & -two fifths & -one fifth \\ one fifth & eight fifths & -six fifths \end{bmatrix}$- 1 5 - 3 5 6 5 1 5 - 2 5 - 1 5 1 5 8 5
3.3.15
Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix. $A = \begin{bmatrix} 2 & 0 & 0 \\ -3 & 1 & 1 \\ -3 & 3 & 1 \end{bmatrix}$
Answer
A. The inverse matrix is $A^{-1}$ $= \begin{bmatrix} one half & 0 & 0 \\ 0 & -one half & one half \\ three halves & three halves & -one half \end{bmatrix}$ - 1 2 1 2 3 2 3 2 - 1 2. (Type an integer or simplified fraction for each matrix ele
3.3.19
Find the area of the parallelogram whose vertices are listed. (0, 0), (3, 7), (8, 4), (11, 11)
Answer
The area of the parallelogram is 44 square units
3.3.21
Find the area of the parallelogram whose vertices are listed. ($- 3, -2$), (0, 5), (4, -4), (7, 3)
Answer
The area of the parallelogram is 55 square units
3.3.23
Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (2, 0, -5), (1, 3, 3), and (6, 1, 0)
Answer
The volume of the parallelepiped is 79. (Type an integer or a decimal.)
§4.1 Vector Spaces and Subspaces
4.1.3
Let $H = \{ \begin{bmatrix} x \\ y \end{bmatrix}$: 6 x squared $+ 4 y$ squared less than or $= 1\}$, which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples long dash two vectors, and a vector and a scalar long dash to show that H is not a subspace of $\mathbb{R}^{2}$
Answer
H is not a subspace of $\mathbb{R}^{2}$ because the two vectors$\begin{bmatrix} \frac{1}{\sqrt{6}} \\ 0 \end{bmatrix}$,$\begin{bmatrix} \frac{1}{\sqrt{6}} \\ 0 \end{bmatrix}$show that H is not closed under addition. (Use a comma to separate vectors as needed.) Part 2 H is not a subspace of $\mathbb{R}^{2}$ because the scalar 2 and the vector$\begin{bmatrix} 0 \\ one half \end{bmatrix}$show that H is not closed under scalar multiplication
4.1.5
Determine if the given set is a subspace of $P_{4}$. Justify your answer. The set of all polynomials of the form $p (t) = a t^{4}$, where a is in R
Answer
A. The set is a subspace of $P_{4}$. The set contains the zero vector of $P_{4}$, the set is closed under vector addition, and the set is closed under multiplication by scalars.
4.1.7
Determine if the given set is a subspace of $P_{8}$. Justify your answer. All polynomials of degree at most 8, with positive real numbers as coefficients
Answer
Complete each statement below. The zero vector of $P_{8}$ is not in the set because zero is not a positive real number. Part 2 The set is closed under vector addition because the sum of two positive real numbers is a positive real number. Part 3 The set is not closed under multiplication by scalars because the product of a scalar and a positive real number is not necessarily a positive real number. Part 4 Is the set a subspace of $P_{8}$ ? Your answer is correct
4.1.8
Determine if the given set is a subspace of $P_{n}$. Justify your answer. The set of all polynomials in $P_{n}$ such that $p (0) = 0$
Answer
D. The set is a subspace of $P_{n}$ because the set contains the zero vector of $P_{n}$, the set is closed under vector addition, and the set is closed under multiplication by scalars.
4.1.9
Let H be the set of all vectors of the form$\begin{bmatrix} -4 t \\ 5 t \\ t \end{bmatrix}$. Find a vector v in $\mathbb{R}^{3}$ such that $H =$ Span $\{ \mathbf{v} \}$. Why does this show that H is a subspace of $\mathbb{R}^{3}$ ?
Answer
C. Since v is in $\mathbb{R}^{3}$, $H =$ Span $\{ \mathbf{v} \}$ } is a subspace of $\mathbb{R}^{3}$.
4.1.10
Let H be the set of all vectors of the form$\begin{bmatrix} 3 t \\ 0 \\ -7 t \end{bmatrix}$. Show that H is a subspace of $\mathbb{R}^{3}$
Answer
A. For any set of vectors in $\mathbb{R}^{3}$, the span of those vectors is a subspace of $\mathbb{R}^{3}$.
4.1.11
Let W be the set of all vectors of the form shown on the right, where b and c are arbitrary. Find vectors u and v such that $W =$ Span $\{ \mathbf{u}$,$\mathbf{v} \}$. Why does this show that W is a subspace of $\mathbb{R}^{3}$ ?$\begin{bmatrix} 4 b - 9 c \\ -b \\ 5 c \end{bmatrix}$
Answer
B. If$\mathbf{v}_{1}$,…, $\mathbf{v}_{p}$ are in a vector space V, then Span $\{ \mathbf{v}_{1}$,…, $\mathbf{v}_{p}\}$ Spanv1,…, vp is a subspace of V.
4.1.12
Let W be the set of all vectors of the form$\begin{bmatrix} 2 s \\ 2 s + 4 t \\ 4 s - 4 t \\ 5 t \end{bmatrix}$. Show that W is a subspace of $\mathbb{R}^{4}$ by finding vectors u and v such that $W =$ Span{ u, v }
Answer
D. $W =$ Span $\{ \mathbf{u}$,$\mathbf{v} \}$ }
4.1.13
Let $v_{1} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$, $v_{2} = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$, $v_{3} = \begin{bmatrix} 7 \\ 3 \\ 14 \end{bmatrix}$, and $w = \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix}$. a. Is w in {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$}? How many vectors are in {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$}? b. How many vectors are in Span{$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$}? c. Is w in the subspace spanned by {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$}? Why?
Answer
B. Vector w is not in {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$} because it is not$\mathbf{v}_{1}$,$\mathbf{v}_{2}$, or$\mathbf{v}_{3}$.
4.1.14
Let $v_{1} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$, $v_{2} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$, $v_{3} = \begin{bmatrix} 8 \\ 2 \\ 2 \end{bmatrix}$, and $w = \begin{bmatrix} 3 \\ 1 \\ 6 \end{bmatrix}$. Is w in the subspace spanned by {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$}? Why?
Answer
A. Vector w is not in the subspace spanned by {$\mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3}$} because the equation $x_{1} \mathbf{v}_{1} + x_{2} \mathbf{v}_{2} + x_{3} \mathbf{v}_{3}$=$\mathbf{w}$ has no solution which can be seen because an echelon form of the augmented matrix of the system has a row of the form [0 $\times$ • $\times$ • $\times$ • 0 b] with b not = ≠0.
4.1.15
Let W be the set of all vectors of the form shown on the right, where a and b represent arbitrary real numbers. Find a set S of vectors that spans W, or give an example or an explanation showing why W is not a vector space.$\begin{bmatrix} 2 a + 3 b \\ -1 \\ 2 a - 7 b \end{bmatrix}$
Answer
B. W is not a vector space because the zero vector and most sums and scalar multiples of vectors in W are not in W, because their second (middle) value $\ne -1$.
4.1.17
Let W be the set of all vectors of the form shown on the right, where a, b, and c represent arbitrary real numbers. Find a set S of vectors that spans W or give an example or an explanation to show that W is not a vector space.$\begin{bmatrix} 2 a + 4 b \\ 9 b - 6 c \\ 6 c - 7 a \\ 7 b \end{bmatrix}$
Answer
A. A spanning set is $S = \{\square\} \begin{bmatrix} 2 \\ 0 \\ -7 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 4 \\ 9 \\ 0 \\ 7 \end{bmatrix}$,
4.1.22
For fixed positive integers m and n, the set Upper $M_{m} \times n$ of all $m \times n$ matrices is a vector space, under the usual operations of addition of matrices and multiplication by real scalars. Let F be a fixed $3 \times 2$ matrix, and let H be the set of all matrices A in Upper $M_{2} \times 4$ with the property that FA $= 0$ (the zero matrix in Upper $M_{3} \times 4$). Determine if H is a subspace of Upper $M_{2} \times 4$
Answer
C. The set H is a subspace of Upper $M_{2} \times 4$ because the set contains the $2 \times$ ×4 zero matrix, the set is closed under addition, and the set is closed under multiplication by scalars.
4.1.24
Determine whether the statement is True or False. Justify your answer. A vector is any element of a vector space
Answer
D. The statement is true. The elements of a vector space are called vectors.
4.1.26
Determine whether the statement is True or False. Justify your answer. If u is a vector in a vector space V, then ($- 1) u$ is the same as the negative of u
Answer
D. The statement is true. For each u in V, there is a vector - -u in V such that - -u =(- -1)u.
4.1.30
Determine whether the statement is True or False. Justify your answer. $\mathbb{R}^{2}$ is a subspace of $\mathbb{R}^{3}$
Answer
D. The statement is false. $\mathbb{R}^{2}$ is not even a subset of $\mathbb{R}^{3}$.
4.1.34
The axioms for a vector space V can be used to prove the elementary properties for a vector space. Because of Axiom 2, Axioms 2 and 4 imply, respectively, that$\mathbf{0}$+$\mathbf{u}$=$\mathbf{u}$and -$\mathbf{u}$+$\mathbf{u}$=$\mathbf{0}$for all$\mathbf{u}$. Complete the proof that $- u$ is unique by showing that if $u + w = 0$, then $w = - u$. Use the ten axioms of a vector space to justify each step. Axioms In the following axioms,$\mathbf{u}$,$\mathbf{v}$, and$\mathbf{w}$are in vector space V and c and d are scalars. 1. The sum$\mathbf{u}$+$\mathbf{v}$is in V. 2.$\mathbf{u}$+$\mathbf{v}$=$\mathbf{v}$+$\mathbf{u}_{3}$. ($\mathbf{u}$+$\mathbf{v}$) +$\mathbf{w}$=$\mathbf{u}$+ ($\mathbf{v}$+$\mathbf{w}$) 4. V has a vector$\mathbf{0}$such that$\mathbf{u}$+$\mathbf{0}$=$\mathbf{u}$. 5. For each$\mathbf{u}$in V, there is a vector -$\mathbf{u}$in V such that$\mathbf{u}$+ ($- 1) \mathbf{u}$=$\mathbf{0}$. 6. The scalar multiple c$\mathbf{u}$is in V. 7. c($\mathbf{u}$+$\mathbf{v}$) $= c \mathbf{u} + c \mathbf{v}_{8}$. $(c + d) \mathbf{u} = c \mathbf{u} + d \mathbf{u}_{9}$. c(d$\mathbf{u}$) = (cd)$\mathbf{u}_{10}$. 1$\mathbf{u}$=$\mathbf{u}$
Answer
Suppose that w satisfies $u + w = 0$. Adding $- u$ to both sides results in the following. Part 2 Part 3 ($- u$) + [ $u + w$ ] = ($- u$) $+ 0$ [($- u$) $+ u$ ] $+ w$ ModifyingAbove = With font size decreased by 7 ($- u$) $+ 0$ by Axiom $3 0 + w$ ModifyingAbove = With font size decreased by 7 ($- u$) $+ 0$ by Axiom 5 w ModifyingAbove = With font size decreased by $7 - u$ by Axiom 4
§4.2 Null Spaces, Column Spaces, Row Spaces, and Linear Transformations
4.2.1
Determine if $w = \begin{bmatrix} 5 \\ 5 \\ -1 \end{bmatrix}$is in Nul A, where $A = \begin{bmatrix} 4 & -2 & -3 \\ 6 & -2 & -5 \\ -3 & 1 & 2 \end{bmatrix}$
Answer
B. No, because Aw $= \begin{bmatrix} 13 \\ 25 \\ -12 \end{bmatrix}$
4.2.3
Find an explicit description of Nul A by listing vectors that span the null space. $A = \begin{bmatrix} 1 & 3 & 4 & 0 \\ 0 & 1 & 2 & -4 \end{bmatrix}$
Answer
A spanning set for Nul A is $\{\square\} \begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -12 \\ 4 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.2.5
Find an explicit description of Nul A by listing vectors that span the null space. $A = \begin{bmatrix} 1 & 4 & 0 & -5 & 0 \\ 0 & 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 0 & -4 \end{bmatrix}$
Answer
A spanning set for Nul A is $\{\square\} \begin{bmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 5 \\ 0 \\ -5 \\ 1 \\ 0 \end{bmatrix}$. (Use a comma to separate answers as needed.)
4.2.6
Find an explicit description of Nul A by listing vectors that span the null space. $A = \begin{bmatrix} 1 & 6 & -3 & -7 & 1 \\ 0 & 1 & -8 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
A spanning set for Nul A is $\{\square\} \begin{bmatrix} -45 \\ 8 \\ 1 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 13 \\ -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.2.8
Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. $W = \{ \begin{bmatrix} r \\ s \\ t \end{bmatrix}$: $4 r - 1 = s + 8 t\}$
Answer
A. The set W would be a subspace of $\mathbb{R}^{3}$.
4.2.9
Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. $W = \{ \begin{bmatrix} p \\ q \\ r \\ s \end{bmatrix}$:$\begin{bmatrix} 3 p - 3 q = -s \\ 2 p = -2 s - 3 r \end{bmatrix} \}$
Answer
B. The set of all solutions to the homogeneous system of equations.
4.2.11
Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. $W = \{ \begin{bmatrix} s - 2 t \\ 3 + 3 s \\ 2 s + t \\ 2 s \end{bmatrix}$: $s, t real\}$
Answer
A. The set W would be a subspace of $\mathbb{R}^{4}$.
4.2.13
Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. $W = \{ \begin{bmatrix} c - 2 d \\ d \\ c \end{bmatrix}$: $c, d real\}$
Answer
Let $w = \begin{bmatrix} c - 2 d \\ d \\ c \end{bmatrix}$. Then w is an element of W. Write w in parametric vector form. $w = c \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + d \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$(c, d in R) Part 2 Now write the expression found for w in the previous step as the product of a matrix, A, and the vector$\begin{bmatrix} c \\ d \end{bmatrix}$. $w =$ A$\begin{bmatrix} c \\ d \end{bmatrix} w = \begin{bmatrix} 1 & -2 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} c \\ d \end{bmatrix}$Part 3 Therefore, the set $W =$ Col A. Part 4 The column space of an $m \times n$ matrix A is a subspace of $\mathbb{R}^{m}$. Part 5 The proof is complete since W is a subspace of $\mathbb{R}^{3}$. The given set W must be a vector space because a subspace itself is a vector space
4.2.15
Find A such that the given set is Col A. $\{ \begin{bmatrix} 3 s + t \\ 3 r - s \\ r - 2 s + 3 t \\ -r + s + 2 t \end{bmatrix}$: $r, s, t real\}$
Answer
A. $A = \begin{bmatrix} 0 & 3 & 1 \\ 3 & -1 & 0 \\ 1 & -2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$A equals
4.2.17
Complete parts (a) and (b) for the matrix below. $A = \begin{bmatrix} -5 & -8 & -5 & -4 \\ -1 & 9 & -8 & 1 \\ 9 & 4 & 3 & -5 \end{bmatrix}$
Answer
(a) Find k such that Nul(A) is a subspace of $\mathbb{R}^{k}$. $k = 4$ Part 2 (b) Find k such that Col(A) is a subspace of $\mathbb{R}^{k}$. $k = 3$
4.2.20
Complete parts (a) and (b) for the matrix below. $A = \begin{bmatrix} -9 & 0 & 9 \end{bmatrix}$
Answer
a. Find k such that Nul(A) is a subspace of $\mathbb{R}^{k}$. $k = 3$ Part 2 b. Find k such that Col(A) is a subspace of $\mathbb{R}^{k}$. $k = 1$
4.2.21
For the matrix A below, find a nonzero vector in Nul A, a nonzero vector in Col A, and a nonzero vector in Row A. $A = \begin{bmatrix} -15 & 6 \\ -5 & 2 \\ 10 & -4 \\ 15 & -6 \end{bmatrix}$
Answer
A nonzero column vector in Nul A is$\begin{bmatrix} 2 \\ 5 \end{bmatrix}$. Part 2 A nonzero column vector in Col A is$\begin{bmatrix} -15 \\ -5 \\ 10 \\ 15 \end{bmatrix}$. Part 3 A nonzero column vector in Row A is$\begin{bmatrix} -5 \\ 2 \end{bmatrix}$
4.2.23
Let $A = \begin{bmatrix} -12 & 36 \\ -4 & 12 \end{bmatrix}$and $w = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$. Determine if w is in Col(A). Is w in Nul(A)?
Answer
B. The vector w is in Col(A) because Ax =w is a consistent system. One solution is $x = \begin{bmatrix} -one fourth \\ 0 \end{bmatrix}$- 1 4 0.
4.2.24
Let $A = \begin{bmatrix} -8 & -2 & -20 \\ 8 & 6 & 28 \\ 2 & 0 & 4 \end{bmatrix}$and $w = \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}$. Determine if w is in Col(A). Is w in Nul(A)?
Answer
D. The vector w is in Col(A) because Ax =w is a consistent system.
4.2.26
A denotes an $m \times n$ matrix. Determine whether the statement is true or false. Justify your answer. A null space is a vector space
Answer
A. The statement is true. The null space of an $m \times$ ×n matrix A is a subspace of $\mathbb{R}^{n}$.
§4.3 Linearly Independent Sets; Bases
4.3.1
Describe the set $\{ \begin{bmatrix} 6 \\ 6 \\ 6 \end{bmatrix}$,$\begin{bmatrix} 6 \\ 6 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 6 \\ 0 \\ 0 \end{bmatrix} \}$
Answer
A. The set is a basis for $\mathbb{R}^{3}$.
4.3.2
Describe the set $\{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \}$
Answer
D. None of the above are true.
4.3.3
Describe the set $\{ \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}$,$\begin{bmatrix} -3 \\ 1 \\ 11 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -1 \\ -8 \end{bmatrix} \}$
Answer
D. None of the above
4.3.4
Describe the set $\{ \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -4 \\ 2 \end{bmatrix}$,$\begin{bmatrix} -8 \\ 4 \\ 4 \end{bmatrix} \}$
Answer
A. The set is linearly independent.
4.3.6
Describe the set $\{ \begin{bmatrix} 1 \\ 3 \\ -9 \end{bmatrix}$,$\begin{bmatrix} -3 \\ 4 \\ 12 \end{bmatrix} \}$
Answer
A. The set is linearly independent.
4.3.8
Describe the set $\{ \begin{bmatrix} 1 \\ -7 \\ 2 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -23 \\ 8 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 3 \\ -3 \end{bmatrix} \}$
Answer
C. The set spans $\mathbb{R}^{3}$.
4.3.9
Find a basis for the null space of the matrix$\begin{bmatrix} 1 & 0 & -7 & 4 \\ 0 & 1 & -5 & 5 \\ 7 & -10 & 1 & -22 \end{bmatrix}$
Answer
A basis for the null space is $\{\square\} \begin{bmatrix} 7 \\ 5 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -4 \\ -5 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.3.10
Find a basis for the null space of the matrix given below.$\begin{bmatrix} 1 & 1 & -3 & 1 & 3 \\ 0 & 1 & 0 & -1 & -1 \\ 0 & 0 & -6 & 0 & 6 \end{bmatrix}$
Answer
A basis for the null space is $\{\square\} \begin{bmatrix} -1 \\ 1 \\ 1 \\ 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$. (Use a comma to separate answers as needed.)
4.3.11
Find a basis for the set of vectors in $\mathbb{R}^{3}$ in the plane $x - 7 y + 4 z = 0$. [Hint: Think of the equation as a “system” of homogeneous equations.]
Answer
A basis for the set of vectors in $\mathbb{R}^{3}$ in the plane $x - 7 y + 4 z = 0$ is $\{\square\} \begin{bmatrix} 7 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -4 \\ 0 \\ 1 \end{bmatrix}$
4.3.12
Find a basis for the set of vectors in $\mathbb{R}^{2}$ on the line $y = 24 x$
Answer
A basis for the set of vectors in $\mathbb{R}^{2}$ on the line $y = 24 x$ is $\{\square\} \begin{bmatrix} 1 \\ 24 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.3.13
Assume that A is row equivalent to B. Find bases for Nul A, Col A, and Row A. $A = \begin{bmatrix} -2 & 6 & -2 & -6 \\ 2 & -9 & -4 & 2 \\ -3 & 12 & 3 & -5 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 0 & 7 & 7 \\ 0 & 3 & 6 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
A column vector basis for Nul A is $\{\square\} \begin{bmatrix} -21 \\ -6 \\ 3 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -21 \\ -4 \\ 0 \\ 3 \end{bmatrix}$. (Use a comma to separate vectors as needed.) Part 2 A column vector basis for Col A is $\{\square\} \begin{bmatrix} -2 \\ 2 \\ -3 \end{bmatrix}$,$\begin{bmatrix} 6 \\ -9 \\ 12 \end{bmatrix}$. (Use a comma to separate vectors as needed.) Part 3 A row vector basis for Row A is $\{\square\} \begin{bmatrix} 1 & 0 & 7 & 7 \end{bmatrix}$,$\begin{bmatrix} 0 & 3 & 6 & 4 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.3.14
Assume that A is row equivalent to B. Find bases for Nul A, Col A, and Row A. $A = \begin{bmatrix} 1 & 1 & 3 & -5 & 8 \\ 1 & 1 & 0 & 1 & 8 \\ 2 & 2 & -5 & 12 & 5 \\ 3 & 3 & 0 & 3 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 1 & 0 & 1 & 3 \\ 0 & 0 & 3 & -6 & 3 \\ 0 & 0 & 0 & 0 & -4 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
A column vector basis for Nul A is $\{\square\} \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \\ 0 \end{bmatrix}$. (Use a comma to separate vectors as needed.) Part 2 A column vector basis for Col A is $\{\square\} \begin{bmatrix} 1 \\ 1 \\ 2 \\ 3 \end{bmatrix}$,$\begin{bmatrix} 3 \\ 0 \\ -5 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 8 \\ 8 \\ 5 \\ 5 \end{bmatrix}$. (Use a comma to separate vectors as needed.) Part 3 A row vector basis for Row A is $\{\square\} \begin{bmatrix} 1 & 1 & 0 & 1 & 3 \end{bmatrix}$,$\begin{bmatrix} 0 & 0 & 3 & -6 & 3 \end{bmatrix}$,$\begin{bmatrix} 0 & 0 & 0 & 0 & -4 \end{bmatrix}$. (Use a comma to separate vectors as needed.)
4.3.16
Find a basis for the space spanned by the given vectors.$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} -6 \\ 0 \\ 0 \\ 6 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -3 \\ 3 \\ -1 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -12 \\ 12 \\ -1 \end{bmatrix}$,$\begin{bmatrix} -1 \\ -3 \\ 3 \\ 2 \end{bmatrix}$
Answer
A basis for the space spanned by the given vectors is $\{\square\} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} -6 \\ 0 \\ 0 \\ 6 \end{bmatrix}$,$\begin{bmatrix} 2 \\ -3 \\ 3 \\ -1 \end{bmatrix}$. (Use a comma to separate answers as needed.)
4.3.19
Let$\mathbf{v}_{1}$=$\begin{bmatrix} 1 \\ -6 \\ 7 \end{bmatrix}$,$\mathbf{v}_{2}$=$\begin{bmatrix} 4 \\ 9 \\ -8 \end{bmatrix}$,$\mathbf{v}_{3}$=$\begin{bmatrix} 9 \\ 1 \\ 3 \end{bmatrix}$, and $H =$ Span $\{ \mathbf{v}_{1}$,$\mathbf{v}_{2}$,$\mathbf{v}_{3} \}$. It can be verified that 7$\mathbf{v}_{1} + 5 \mathbf{v}_{2} - 3 \mathbf{v}_{3} = 0$. Use this information to find a basis for H
Answer
A basis for H is $\{\square\} \begin{bmatrix} 1 \\ -6 \\ 7 \end{bmatrix}$,$\begin{bmatrix} 4 \\ 9 \\ -8 \end{bmatrix}$. (Type an integer or decimal for each matrix element. Use a comma to separate vectors as needed.)
4.3.22
Determine if the following statement is true or false. Justify the answer. A linearly independent set in a subspace H is a basis for H
Answer
C. The statement is false because the subspace spanned by the set must also coincide with H.
4.3.24
Determine if the following statement is true or false. Justify the answer. If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V
Answer
A. The statement is true by the Spanning Set Theorem.
4.3.26
Determine if the following statement is true or false. Justify the answer. A basis is a linearly independent set that is as large as possible
Answer
B. The statement is true by the definition of a basis.
4.3.28
Determine if the following statement is true or false. Justify the answer. The standard method for producing a spanning set for Nul A sometimes fails to produce a basis for Nul A
Answer
C. The statement is false because the method always produces an independent set.
4.3.30
Determine if the following statement is true or false. Justify the answer. If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A
Answer
A. Choose the correct answer below. Your answer is correct
4.3.32
Determine if the following statement is true or false. Justify the answer. If A and B are row equivalent, then their row spaces are the same
Answer
A. The statement is true. If B is obtained from A by row operations, the rows of B are linear combinations of the rows of A and vice-versa.
4.3.36
In the vector space of all real-valued functions, find a basis for the subspace spanned by $\{sine t$, sine 2 t, sine t cosine $t\}$
Answer
A basis for this subspace is $\{sine t$, sine $2 t\}$
4.3.39
Let $S = \{ \mathbf{v}_{1}$,…, $\mathbf{v}_{k}\}$ be a set of k vectors in $\mathbb{R}^{n}$, with k less than n. Use a theorem
Answer
A. The matrix A has a pivot position in each row.
4.3.44
Consider the polynomials$\mathbf{p}_{1} (t) = 4 + 3 t$,$\mathbf{p}_{2} (t) = 4 - 3 t$, and$\mathbf{p}_{3} (t) = 8$ (for all t). By inspection, write a linear dependence relation among$\mathbf{p}_{1}$,$\mathbf{p}_{2}$, and$\mathbf{p}_{3}$. Then find a basis for Span $\{ \mathbf{p}_{1}$,$\mathbf{p}_{2}$,$\mathbf{p}_{3} \}$
Answer
B. {$\mathbf{p}_{1}$,$\mathbf{p}_{2}$}
§4.5 The Dimension of a Vector Space
4.5.6
For the subspace below, (a) find a basis, and (b) state the dimension. $\{ \begin{bmatrix} 6 a + 12 b - 2 c \\ 9 a - 3 b - 3 c \\ -12 a + 5 b + 4 c \\ -3 a + b + c \end{bmatrix}$: a, b, c in $R\}$
Answer
a. Find a basis for the subspace. A basis for the subspace is $\{\square\} \begin{bmatrix} 12 \\ -3 \\ 5 \\ 1 \end{bmatrix}$,$\begin{bmatrix} -2 \\ -3 \\ 4 \\ 1 \end{bmatrix}$. (Use a comma to separate vectors as needed.) Part 2 b. State the dimension. The dimension is 2
4.5.8
For the subspace below, (a) find a basis for the subspace, and (b) state the dimension. $\{(a, b, c, d)$ : $a - 5 b + 6 c = 0\}$
Answer
(a) Find a basis for the subspace. A basis for the subspace is $\{\square\} \begin{bmatrix} 5 \\ 1 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -6 \\ 0 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate matrices as needed.) Part 2 (b) State the dimension. The dimension is 3
4.5.9
Find the dimension of the subspace spanned by the given vectors.$\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$,$\begin{bmatrix} 2 \\ 1 \\ -3 \end{bmatrix}$,$\begin{bmatrix} 5 \\ 4 \\ -6 \end{bmatrix}$,$\begin{bmatrix} -4 \\ -3 \\ 5 \end{bmatrix}$
Answer
The dimension of the subspace spanned by the given vectors is 2
4.5.10
Find the dimension of the subspace spanned by the given vectors.$\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 22 \\ 71 \\ 5 \end{bmatrix}$,$\begin{bmatrix} 3 \\ 2 \\ 8 \end{bmatrix}$
Answer
The dimension of the subspace spanned by the given vectors is 3
4.5.11
Determine the dimensions of Nul A, Col A, and Row A for the given matrix. $A = \begin{bmatrix} 1 & 8 & -2 & -6 & 1 \\ 0 & 1 & -1 & -2 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
The dimension of Nul A is 3. (Type a whole number.) Part 2 The dimension of Col A is 2. (Type a whole number.) Part 3 The dimension of Row A is 2. (Type a whole number.)
4.5.12
Determine the dimensions of Nul A, Col A, and Row A for the given matrix. $A = \begin{bmatrix} 1 & 4 & -6 & 2 & -3 & 6 & 0 \\ 0 & 0 & 1 & -4 & 5 & 3 & -5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Answer
The dimension of Nul A is 5. (Type a whole number.) Part 2 The dimension of Col A is 2. (Type a whole number.) Part 3 The dimension of Row A is 2. (Type a whole number.)
4.5.13
Determine the dimensions of Nul A, Col A, and Row A for the given matrix. $A = \begin{bmatrix} 1 & 4 & 0 & 5 \\ 0 & 1 & 0 & -2 \end{bmatrix}$
Answer
The dimension of Nul A is 2. (Type a whole number.) Part 2 The dimension of Col A is 2. (Type a whole number.) Part 3 The dimension of Row A is 2. (Type a whole number.)
4.5.15
Determine the dimensions of Nul A, Col A, and Row A for the given matrix. $A = \begin{bmatrix} 1 & -5 & 0 \\ 0 & 1 & 7 \\ 0 & 0 & 1 \end{bmatrix}$
Answer
The dimension of Nul A is 0. (Type a whole number.) Part 2 The dimension of Col A is 3. (Type a whole number.) Part 3 The dimension of Row A is 3. (Type a whole number.)
4.5.18
Let A be an $m \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. The number of variables in the equation $A x = 0 =$ the nullity of A
Answer
A. Choose the correct answer below. Your answer is correct
4.5.23
Let A be an $m \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A
Answer
A. Choose the correct answer below. Your answer is correct
4.5.24
Let A be an $m \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. The nullity of A is the number of columns of A that are not pivot columns
Answer
B. The statement is true. The nullity of $A =$ the number of free variables in the equation Ax =0.
4.5.37
If the nullity of $a_{4} \times 6$ matrix A is 3, what are the dimensions of the column and row spaces of A?
Answer
dim Col $A = 3$ (Simplify your answer.) Part 2 dim Row $A = 3$ (Simplify your answer.)
§5.1 Eigenvectors and Eigenvalues
5.1.2
Is lambda $= -4$ an eigenvalue of $A = \begin{bmatrix} -9 & 5 \\ -2 & -2 \end{bmatrix}$? Why or why not?
Answer
B. Yes, lambda λ is an eigenvalue of A because Ax $= $ lambda λx has a nontrivial solution.
5.1.4
Is $v = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$an eigenvector of $A = \begin{bmatrix} -1 & 4 \\ 3 & 3 \end{bmatrix}$? If so, find the eigenvalue
Answer
A. The eigenvalue is lambda λ $= -3$.
5.1.6
Is $v = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$an eigenvector of $A = \begin{bmatrix} -4 & 3 & 3 \\ 2 & -3 & -2 \\ -1 & 0 & -2 \end{bmatrix}$? If so, find the eigenvalue
Answer
A. The eigenvalue is lambda λ $= -1$.
5.1.7
Is lambda $= 8$ an eigenvalue of$\begin{bmatrix} 7 & 0 & -3 \\ 2 & 7 & 5 \\ -3 & 4 & 3 \end{bmatrix}$? If so, find one corresponding eigenvector
Answer
A. Yes, lambda λ $= 8 8$ is an eigenvalue of$\begin{bmatrix} 7 & 0 & -3 \\ 2 & 7 & 5 \\ -3 & 4 & 3 \end{bmatrix}$. One corresponding eigenvector is$\begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix}$
5.1.8
Is lambda $= 8$ an eigenvalue of$\begin{bmatrix} 6 & 2 & 2 \\ 2 & 3 & 4 \\ 0 & 1 & 6 \end{bmatrix}$? If so, find one corresponding eigenvector
Answer
A. Yes, lambda λ $= 8 8$ is an eigenvalue of$\begin{bmatrix} 6 & 2 & 2 \\ 2 & 3 & 4 \\ 0 & 1 & 6 \end{bmatrix}$. One corresponding eigenvector is$\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$. (Type a vector or list of vector
5.1.9
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below. $A = \begin{bmatrix} 1 & 0 \\ -2 & 3 \end{bmatrix}$, lambda $= 3, 1$
Answer
A basis for the eigenspace corresponding to lambda $= 3$ is $\{\square\} \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. (Use a comma to separate answers as needed.) Part 2 A basis for the eigenspace corresponding to lambda $= 1$ is $\{\square\} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. (Use a comma to separate answers as needed.)
5.1.11
Find a basis for the eigenspace corresponding to the eigenvalue. $A = \begin{bmatrix} 2 & -6 \\ -4 & 7 \end{bmatrix}$, lambda $= 10$
Answer
A basis for the eigenspace corresponding to lambda $= 10$ is $\{\square\} \begin{bmatrix} -3 \\ 4 \end{bmatrix}$. (Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)
5.1.12
Find a basis for the eigenspace corresponding to each listed eigenvalue. $A = \begin{bmatrix} 7 & 4 \\ -3 & -1 \end{bmatrix}$, lambda $= 1, 5$
Answer
A basis for the eigenspace corresponding to lambda $= 1$ is $\{\square\} \begin{bmatrix} -2 \\ 3 \end{bmatrix}$. (Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.) Part 2 A basis for the eigenspace corresponding to lambda $= 5$ is $\{\square\} \begin{bmatrix} -2 \\ 1 \end{bmatrix}$. (Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)
5.1.13
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below. $A = \begin{bmatrix} 3 & 1 & 0 \\ -2 & 0 & 0 \\ -4 & 2 & 5 \end{bmatrix}$, lambda $= 5, 1, 2$
Answer
A basis for the eigenspace corresponding to lambda $= 5$ is $\{\square\} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate answers as needed.) Part 2 A basis for the eigenspace corresponding to lambda $= 1$ is $\{\square\} \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}$. (Use a comma to separate answers as needed.) Part 3 A basis for the eigenspace corresponding to lambda $= 2$ is $\{\square\} \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$. (Use a comma to separate answers as needed.)
5.1.14
Find a basis for the eigenspace corresponding to the eigenvalue of A given below. $A = \begin{bmatrix} 6 & 0 & -2 \\ 3 & 0 & -16 \\ -2 & 1 & 11 \end{bmatrix}$, lambda $= 5$
Answer
A basis for the eigenspace corresponding to lambda $= 5$ is $\{\square\} \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}$. (Use a comma to separate answers as needed.)
5.1.15
Find a basis for the eigenspace corresponding to the eigenvalue. $A = \begin{bmatrix} 4 & 1 & 1 \\ 4 & 7 & 4 \\ 3 & 3 & 6 \end{bmatrix}$, lambda $= 3$
Answer
A basis for the eigenspace corresponding to lambda $= 3$ is $\{\square\} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$. (Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)
5.1.16
Find a basis for the eigenspace corresponding to the eigenvalue of A given below. $A = \begin{bmatrix} 4 & 0 & 2 & 0 \\ 3 & 2 & 5 & 0 \\ 4 & -3 & 8 & 0 \\ 5 & -2 & 8 & 3 \end{bmatrix}$, lambda $= 3$
Answer
A basis for the eigenspace corresponding to lambda $= 3$ is $\{\square\} \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$. (Use a comma to separate answers as needed.)
5.1.17
Find the eigenvalues of the matrix.$\begin{bmatrix} 0 & 0 & 0 \\ 0 & -8 & 9 \\ 0 & 0 & -4 \end{bmatrix}$
Answer
The eigenvalue(s) of the matrix is/are 0, -8, -4. (Use a comma to separate answers as needed.)
5.1.22
A is an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If A$\mathbf{x}$= lambda$\mathbf{x}$for some scalar lambda, then$\mathbf{x}$is an eigenvector of A
Answer
A. The vector x must be nonzero.
§5.2 The Characteristic Equation
5.2.1
Find the characteristic polynomial and the eigenvalues of the matrix.$\begin{bmatrix} 9 & 4 \\ 4 & 9 \end{bmatrix}$
Answer
A. The real eigenvalue(s) of the matrix is/are 5, 13 5, 13. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)
5.2.3
Find the characteristic polynomial and the eigenvalues of the matrix.$\begin{bmatrix} -3 & 6 \\ 1 & -1 \end{bmatrix}$
Answer
A. The real eigenvalue(s) of the matrix is/are $-2 +$ or $- \sqrt{7}$ -2± 7. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)
5.2.4
Find the characteristic polynomial and the eigenvalues of the matrix.$\begin{bmatrix} 6 & -2 \\ -5 & 2 \end{bmatrix}$
Answer
A. The real eigenvalue(s) of the matrix is/are $4 +$ or $- \sqrt{14}$ 4± 14. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)
5.2.5
Find the characteristic polynomial and the eigenvalues of the matrix.$\begin{bmatrix} -9 & 1 \\ -1 & -7 \end{bmatrix}$
Answer
A. The real eigenvalue(s) of the matrix is/are -8. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)
5.2.6
Find the characteristic polynomial and the eigenvalues of the matrix.$\begin{bmatrix} 6 & 2 \\ -2 & 5 \end{bmatrix}$
Answer
B. The matrix has no real eigenvalues.
5.2.9
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} 1 & 0 & -1 \\ 3 & 4 & -2 \\ 0 & 5 & 0 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 5$ lambda squared $- 14$ lambda $- 5$. (Type an expression using lambda as the variable.)
5.2.10
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} 0 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 0 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 14$ lambda $+ 12$. (Type an expression using lambda as the variable.)
5.2.11
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} 3 & 0 & 0 \\ -8 & 5 & -3 \\ -6 & 0 & 2 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 10$ lambda squared $- 31$ lambda $+ 30$. (Type an expression using lambda as the variable.)
5.2.12
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} -2 & 0 & -8 \\ -8 & 5 & -6 \\ 0 & 0 & 6 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 9$ lambda squared $- 8$ lambda $- 60$. (Type an expression using lambda as the variable.)
5.2.13
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} 5 & -3 & 0 \\ -3 & 9 & 0 \\ 5 & 6 & 4 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 18$ lambda squared $- 92$ lambda $+ 144$. (Type an expression using lambda as the variable.)
5.2.14
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants. [Note: Finding the characteristic polynomial of $a_{3} \times 3$ matrix is not easy to do with just row operations, because the variable lambda is involved.]$\begin{bmatrix} 6 & 9 & 3 \\ 0 & 3 & 0 \\ 7 & 6 & -4 \end{bmatrix}$
Answer
The characteristic polynomial is negative lambda cubed $+ 5$ lambda squared $+ 39$ lambda $- 135$. (Type an expression using lambda as the variable.)
5.2.15
For the matrix, list the real eigenvalues, repeated according to their multiplicities.$\begin{bmatrix} 9 & -3 & 0 & 6 \\ 0 & 6 & 2 & -3 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 0 & 9 \end{bmatrix}$
Answer
The real eigenvalues are 3, 6, 9, 9. (Use a comma to separate answers as needed.)
5.2.16
For the matrix, list the eigenvalues, repeated according to their multiplicities.$\begin{bmatrix} 3 & 0 & 0 & 0 \\ -8 & 3 & 0 & 0 \\ 0 & -7 & -8 & 0 \\ 5 & -4 & -7 & 9 \end{bmatrix}$
Answer
The eigenvalues are 3, 3, -8, 9. (Use a comma to separate answers as needed.)
5.2.18
It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find h in the matrix A below such that the eigenspace for lambda $= 5$ is two-dimensional. $A = \begin{bmatrix} 5 & -2 & 8 & 5 \\ 0 & 3 & h & 0 \\ 0 & 0 & 5 & 2 \\ 0 & 0 & 0 & -2 \end{bmatrix}$
Answer
The value of h for which the eigenspace for lambda $= 5$ is two-dimensional is $h = 8$
5.2.27
Let A be an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If lambda $+ 5$ is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A
Answer
A. In order for 5 to be an eigenvalue of A, the characteristic polynomial would need to have a factor of lambda λ - -5.
§5.3 Diagonalization
5.3.11
Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.$\begin{bmatrix} 5 & 7 & -10 \\ 2 & 12 & -12 \\ 2 & 8 & -8 \end{bmatrix}$; lambda $= 2, 3, 4$
Answer
A. For $P = \begin{bmatrix} 1 & 3 & 2 \\ 1 & 2 & 4 \\ 1 & 2 & 3 \end{bmatrix} 1 3 2 1 2 4 1 2 3, D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$
5.3.12
Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.$\begin{bmatrix} 5 & 2 & 2 \\ 2 & 5 & 2 \\ 2 & 2 & 5 \end{bmatrix}$; lambda $= 3, 9$
Answer
A. For $P = \begin{bmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$ $1 1 0 1 0 1 1, D =$
5.3.15
Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.$\begin{bmatrix} -2 & 1 & 1 \\ -4 & 3 & 4 \\ -2 & 2 & 1 \end{bmatrix}$; lambda $= - 1, 4$
Answer
B. For $P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 4 \\ 0 & 1 & 2 \end{bmatrix} 1 1 1 1 0 4 0 1 2, D =$
5.3.16
Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.$\begin{bmatrix} 0 & -4 & -6 \\ -1 & 0 & -3 \\ 1 & 2 & 5 \end{bmatrix}$; lambda $= 1, 2$
Answer
A. For $P = \begin{bmatrix} -2 & -2 & -3 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$ $1 0 1 0 1, D =$
5.3.17
Diagonalize the following matrix.$\begin{bmatrix} 4 & 0 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
Answer
C. The matrix cannot be diagonalized.
5.3.18
Diagonalize the following matrix. One eigenvalue is lambda $= 5$ and one eigenvector is (-4, 3, 2).$\begin{bmatrix} -19 & -36 & 24 \\ 18 & 32 & -18 \\ 12 & 18 & -7 \end{bmatrix}$
Answer
B. For $P = \begin{bmatrix} -4 & -3 & 1 \\ 3 & 2 & 0 \\ 2 & 0 & 1 \end{bmatrix}$ $1 3 2 0 2 0 1, D =$
5.3.19
Diagonalize the following matrix.$\begin{bmatrix} 7 & -5 & 0 & 5 \\ 0 & 3 & 1 & -2 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}$
Answer
A. For $P = \begin{bmatrix} -1 & 1 & 5 & 1 \\ -1 & 2 & 4 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}$ $2 4 0 1 0 0 0 0 1 0 0, D =$
5.3.21
Let A, P, and D be $n \times n$ matrices. Determine whether the statement below is true or false. Justify the answer. A is diagonalizable if $A = PDP^{-1}$ for some matrix D and some invertible matrix P
Answer
D. The statement is false. The symbol D does not automatically denote a diagonal matrix.
5.3.22
Let A be an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If $\mathbb{R}^{n}$ has a basis of eigenvectors of A, then A is diagonalizable
Answer
D. The statement is true. A is diagonalizable if and only if there are enough eigenvectors to form a basis of $\mathbb{R}^{n}$.
5.3.26
Let A be an $n \times n$ matrix. Determine whether the statement below is true or false. Justify the answer. If A is diagonalizable, then A has n distinct eigenvalues
Answer
B. The statement is false. A diagonalizable matrix can have fewer than n eigenvalues and still have n linearly independent eigenvectors.
5.3.27
Assume A, P, and D are $n \times n$ matrices. Determine whether the statement below is true or false. Justify the answer. If AP = PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A
Answer
A. Choose the correct answer below. Your answer is correct
5.3.30
A is $a_{3} \times 3$ matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?
Answer
B. No. The sum of the dimensions of the eigenspaces $= 2 2$ and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.
5.3.1
Let $A = PDP^{-1}$ and P and D as shown below. Compute $A^{4}$. $P = \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix}$, $D = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}$
Answer
$A^{4} = \begin{bmatrix} 641 & -320 \\ 1120 & -559 \end{bmatrix}$(Simplify your answer.)
5.3.2
Let $A = PDP^{-1}$ and P and D as shown below. Compute $A^{4}$. $P = \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix}$, $D = \begin{bmatrix} 1 & 0 \\ 0 & one half \end{bmatrix}$
Answer
$A^{4} = \begin{bmatrix} \frac{151}{16} & \frac{45}{8} \\ -\frac{225}{16} & -\frac{67}{8} \end{bmatrix}$(Simplify your answer.)
5.3.3
Use the factorization $A = PDP^{-1}$ to compute $A^{k}$, where k represents an arbitrary integer.$\begin{bmatrix} a & 5 (b - a) \\ 0 & b \end{bmatrix}$=$\begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} 1 & -5 \\ 0 & 1 \end{bmatrix}$
Answer
$A^{k} = \begin{bmatrix} a^{k} & 5 (b^{k} - a^{k}) \\ 0 & b^{k} \end{bmatrix}$
§6.1 Inner Product, Length, and Orthogonality
6.1.1
Compute$\mathbf{u} \times \mathbf{u}$,$\mathbf{v} \times \mathbf{u}$, and $\frac{ \mathbf{v} \times \mathbf{u}$}{$\mathbf{u} \times \mathbf{u}$} using the vectors$\mathbf{u}$=$\begin{bmatrix} -3 \\ 5 \end{bmatrix}$and$\mathbf{v}$=$\begin{bmatrix} 7 \\ 4 \end{bmatrix}$
Answer
$\mathbf{u} \times \mathbf{u} = 34$ (Simplify your answer.) Part 2$\mathbf{v} \times \mathbf{u} = -1$ (Simplify your answer.) Part $3 \frac{ \mathbf{v} \times \mathbf{u}$}{$\mathbf{u} \times \mathbf{u}$} = negative $\frac{1}{34}$ (Type an integer or a simplified fraction.)
6.1.2
Compute$\mathbf{w} \times \mathbf{w}$,$\mathbf{x} \times \mathbf{w}$, and $\frac{ \mathbf{x} \times \mathbf{w}$}{$\mathbf{w} \times \mathbf{w}$} using the vectors$\mathbf{w}$=$\begin{bmatrix} 3 \\ -4 \\ -5 \end{bmatrix}$and$\mathbf{x}$=$\begin{bmatrix} 5 \\ -3 \\ 1 \end{bmatrix}$
Answer
$\mathbf{w} \times \mathbf{w} = 50$ (Simplify your answer. Type an integer or a simplified fraction.) Part 2$\mathbf{x} \times \mathbf{w} = 22$ (Simplify your answer. Type an integer or a simplified fraction.) Part $3 \frac{ \mathbf{x} \times \mathbf{w}$}{$\mathbf{w} \times \mathbf{w}$} $= \frac{11}{25}$ (Simplify your answer. Type an integer or a simplified fraction.)
6.1.5
Compute the quantity using the vectors$\mathbf{u}$=$\begin{bmatrix} -5 \\ 2 \end{bmatrix}$, and$\mathbf{v}$=$\begin{bmatrix} 4 \\ 5 \end{bmatrix}$. $(\frac{ \mathbf{u} \times \mathbf{v}$}{$\mathbf{v} \times \mathbf{v}$})$\mathbf{v}$
Answer
$(\frac{ \mathbf{u} \times \mathbf{v}$}{$\mathbf{v} \times \mathbf{v}$})$\mathbf{v}$=$\begin{bmatrix} \square$negative $\frac{40}{41}$ negative $\frac{50}{41}$ (Simplify your answers.)
6.1.7
Compute $\lVert \mathbf{w} \rVert$ using$\mathbf{w}$=$\begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix}$
Answer
$\lVert \mathbf{w} \rVert = \sqrt{38}$ (Type an exact answer, using radicals as needed.)
6.1.9
Find a unit vector in the direction of the given vector.$\begin{bmatrix} 27 \\ -36 \end{bmatrix}$
Answer
A unit vector in the direction of the given vector is$\begin{bmatrix} three fifths \\ -four fifths \end{bmatrix}$. (Type an exact answer, using radicals as needed.)
6.1.10
Find a unit vector in the direction of the given vector.$\begin{bmatrix} -3 \\ 27 \\ -3 \end{bmatrix}$
Answer
A unit vector in the direction of the given vector is$\begin{bmatrix} -\frac{1}{\sqrt{83}} \\ \frac{9}{\sqrt{83}} \\ -\frac{1}{\sqrt{83}} \end{bmatrix}$. (Type an exact answer, using radicals as needed.)
6.1.14
Find the distance between $u = \begin{bmatrix} 0 \\ -6 \\ 2 \end{bmatrix}$and $z = \begin{bmatrix} -5 \\ -1 \\ 6 \end{bmatrix}$
Answer
The distance between u and z is $\sqrt{66}$. (Type an exact answer, using radicals as needed.)
6.1.15
Determine if the following vectors are orthogonal. $a = \begin{bmatrix} 8 \\ -7 \end{bmatrix}$, $b = \begin{bmatrix} -2 \\ -5 \end{bmatrix}$
Answer
D. The vectors a and b are not orthogonal because $a \times$ •b $= 19$.
6.1.16
Determine if the following vectors are orthogonal. $u = \begin{bmatrix} 12 \\ 5 \\ 1 \end{bmatrix}$, $v = \begin{bmatrix} 1 \\ -5 \\ 13 \end{bmatrix}$
Answer
B. The vectors u and v are orthogonal because $u \times$ •v $= 0$.
6.1.17
Determine if the following vectors are orthogonal. $u = \begin{bmatrix} 8 \\ 2 \\ -7 \\ 0 \end{bmatrix}$, $v = \begin{bmatrix} -4 \\ 9 \\ -2 \\ 4 \end{bmatrix}$
Answer
B. The vectors u and v are orthogonal because $u \times$ •v $= 0$.
6.1.18
Determine if the following vectors are orthogonal. $u = \begin{bmatrix} -2 \\ 7 \\ 5 \\ 0 \end{bmatrix}$, $v = \begin{bmatrix} 1 \\ -6 \\ 19 \\ -5 \end{bmatrix}$
Answer
B. The vectors u and v are not orthogonal because $u \times$ •v $= 51$.
6.1.19
Determine whether the statement below is true or false. Justify the answer. The vector is in $\mathbb{R}^{n}$.$\mathbf{v} \times \mathbf{v} = \lVert \mathbf{v} \rVert$ squared
Answer
A. The statement is true. By the definition of the length of a vector$\mathbf{v} v, \lVert v \rVert v = \sqrt{ \mathbf{v} \times \mathbf{v}$} v•v.
§6.2 Orthogonal Sets
6.2.2
Determine whether the set of vectors is orthogonal.$\begin{bmatrix} 1 \\ -7 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 1 \\ 7 \end{bmatrix}$,$\begin{bmatrix} -50 \\ -7 \\ 1 \end{bmatrix}$
Answer
A. The set of vectors is orthogonal because each pair of distinct vectors from the set is orthogonal.
6.2.3
Determine whether the set of vectors is orthogonal.$\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}$,$\begin{bmatrix} -5 \\ 2 \\ 1 \end{bmatrix}$
Answer
Is the set of vectors orthogonal? Your answer is correct
6.2.5
Determine whether the set of vectors is orthogonal.$\begin{bmatrix} 6 \\ -3 \\ 1 \\ 6 \end{bmatrix}$,$\begin{bmatrix} -1 \\ 6 \\ -6 \\ 5 \end{bmatrix}$,$\begin{bmatrix} 12 \\ 35 \\ 33 \\ 0 \end{bmatrix}$
Answer
Is the set of vectors orthogonal? Your answer is correct
6.2.6
Determine whether the set of vectors is orthogonal.$\begin{bmatrix} 4 \\ -5 \\ 0 \\ 3 \end{bmatrix}$,$\begin{bmatrix} -5 \\ 2 \\ -3 \\ 10 \end{bmatrix}$,$\begin{bmatrix} 3 \\ 3 \\ 4 \\ 1 \end{bmatrix}$
Answer
Is the set of vectors orthogonal? Your answer is correct
6.2.7
Show that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is an orthogonal basis for $\mathbb{R}^{2}$. Then express x as a linear combination of the u ’s.$\mathbf{u}_{1}$=$\begin{bmatrix} 6 \\ -8 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} 16 \\ 12 \end{bmatrix}$, and $x = \begin{bmatrix} 7 \\ -1 \end{bmatrix}$
Answer
B. The vectors must form an orthogonal set.
6.2.10
Show that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2}$,$\mathbf{u}_{3} \}$ is an orthogonal basis for $\mathbb{R}^{3}$. Then express x as a linear combination of the u ’s.$\mathbf{u}_{1}$=$\begin{bmatrix} 4 \\ -4 \\ 0 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}$,$\mathbf{u}_{3}$=$\begin{bmatrix} 1 \\ 1 \\ 4 \end{bmatrix}$, and $x = \begin{bmatrix} 5 \\ -2 \\ 1 \end{bmatrix}$
Answer
B. The vectors must span W.
6.2.11
Compute the orthogonal projection of$\begin{bmatrix} 1 \\ 7 \end{bmatrix}$onto the line through$\begin{bmatrix} -8 \\ 4 \end{bmatrix}$and the origin
Answer
The orthogonal projection is$\begin{bmatrix} -2 \\ 1 \end{bmatrix}$. (Simplify your answer.)
6.2.13
Let$\mathbf{y}$=$\begin{bmatrix} 1 \\ 4 \end{bmatrix}$and$\mathbf{u}$=$\begin{bmatrix} 7 \\ -6 \end{bmatrix}$. Write$\mathbf{y}$as the sum of two orthogonal vectors, one in Span $\{ \mathbf{u} \}$ and one orthogonal to$\mathbf{u}$
Answer
$\mathbf{y}$= ModifyingAbove$\mathbf{y}$with caret +$\mathbf{z}$=$\begin{bmatrix} -seven fifths \\ six fifths \end{bmatrix}$+$\begin{bmatrix} \frac{12}{5} \\ \frac{14}{5} \end{bmatrix}$(Type an integer or simplified fraction for each matrix element. List the terms in the same order as they appear in the original list.)
6.2.14
Let$\mathbf{y}$=$\begin{bmatrix} 1 \\ 2 \end{bmatrix}$and$\mathbf{u}$=$\begin{bmatrix} 4 \\ 3 \end{bmatrix}$. Write$\mathbf{y}$as the sum of a vector in Span $\{ \mathbf{u} \}$ and a vector orthogonal to$\mathbf{u}$
Answer
$\mathbf{y}$= ModifyingAbove$\mathbf{y}$with caret +$\mathbf{z}$=$\begin{bmatrix} eight fifths \\ six fifths \end{bmatrix}$+$\begin{bmatrix} -three fifths \\ four fifths \end{bmatrix}$(Type an integer or simplified fraction for each matrix element. List the terms in the same order as they appear in the original list.)
6.2.17
Determine whether the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.$\mathbf{u}_{1}$=$\begin{bmatrix} one third \\ two thirds \\ one third \end{bmatrix}$and$\mathbf{u}_{2}$=$\begin{bmatrix} one half \\ 0 \\ -one half \end{bmatrix}$
Answer
A. The set of vectors is orthogonal only. The normalized vectors for$\mathbf{u}_{1}$ and$\mathbf{u}_{2}$ are$\begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{bmatrix}$ and
6.2.19
Determine if the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.$\mathbf{u}$=$\begin{bmatrix} -0.8 \\ -0.6 \end{bmatrix}$,$\mathbf{v}$=$\begin{bmatrix} -0.6 \\ 0.8 \end{bmatrix}$
Answer
B. The set of vectors is orthonormal.
6.2.21
Determine whether the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.$\mathbf{u}_{1}$=$\begin{bmatrix} \frac{1}{\sqrt{226}} \\ \frac{15}{\sqrt{452}} \\ \frac{15}{\sqrt{452}} \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} \frac{15}{\sqrt{226}} \\ -\frac{1}{\sqrt{452}} \\ -\frac{1}{\sqrt{452}} \end{bmatrix}$,$\mathbf{u}_{3}$=$\begin{bmatrix} 0 \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$
Answer
B. The set of vectors is orthonormal.
§6.3 Orthogonal Projections
6.3.4
Verify that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is an orthogonal set, and then find the orthogonal projection of y onto Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$. $y = \begin{bmatrix} 6 \\ 2 \\ -5 \end{bmatrix}$,$\mathbf{u}_{1}$=$\begin{bmatrix} 5 \\ 6 \\ 0 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} -6 \\ 5 \\ 0 \end{bmatrix}$
Answer
To verify that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is an orthogonal set, find$\mathbf{u}_{1} \times \mathbf{u}_{2}$.$\mathbf{u}_{1} \times \mathbf{u}_{2} = 0$ (Simplify your answer.) Part 2 The projection of y onto Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is$\begin{bmatrix} \square \\ \square \\ \square \end{bmatrix}$. (Simplify your answers.)
6.3.5
Verify that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is an orthogonal set, and then find the orthogonal projection of y onto Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$. $y = \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix}$,$\mathbf{u}_{1}$=$\begin{bmatrix} 5 \\ 1 \\ -2 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix}$
Answer
To verify that $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is an orthogonal set, find$\mathbf{u}_{1} \times \mathbf{u}_{2}$.$\mathbf{u}_{1} \times \mathbf{u}_{2} = 0$ (Simplify your answer.) Part 2 The projection of y onto Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ is$\begin{bmatrix} \square \\ \square \\ \square \end{bmatrix}$. (Simplify your answers.)
6.3.8
Let W be the subspace spanned by$\mathbf{u}_{1}$and$\mathbf{u}_{2}$, and write$\mathbf{y}$as the sum of a vector in W and a vector orthogonal to W.$\mathbf{y}$=$\begin{bmatrix} -3 \\ 5 \\ 4 \end{bmatrix}$,$\mathbf{u}_{1}$=$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} -2 \\ 5 \\ -3 \end{bmatrix}$
Answer
The sum is$\mathbf{y}$= ModifyingAbove$\mathbf{y}$with caret +$\mathbf{z}$, where ModifyingAbove$\mathbf{y}$with caret =$\begin{bmatrix} 1 \\ nine halves \\ one half \end{bmatrix}$is in W and$\mathbf{z}$=$\begin{bmatrix} -4 \\ one half \\ seven halves \end{bmatrix}$is orthogonal to W. (Simplify your answers.)
6.3.9
Let W be a subspace spanned by the$\mathbf{u}$’s, and write$\mathbf{y}$as the sum of a vector in W and a vector orthogonal to W.$\mathbf{y}$=$\begin{bmatrix} 2 \\ 5 \\ 5 \\ -1 \end{bmatrix}$,$\mathbf{u}_{1}$=$\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix}$,$\mathbf{u}_{2}$=$\begin{bmatrix} -1 \\ 5 \\ 3 \\ -4 \end{bmatrix}$,$\mathbf{u}_{3}$=$\begin{bmatrix} -1 \\ 0 \\ 1 \\ 1 \end{bmatrix}$
Answer
$\mathbf{y}$=$\begin{bmatrix} \frac{26}{51} \\ \frac{104}{17} \\ \frac{160}{51} \\ -\frac{32}{51} \end{bmatrix}$+$\begin{bmatrix} \frac{76}{51} \\ -\frac{19}{17} \\ \frac{95}{51} \\ -\frac{19}{51} \end{bmatrix}$(Type an integer or simplified fraction for each matrix element.)
6.3.21
Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in $\mathbb{R}^{n}$. If$\mathbf{z}$is orthogonal to$\mathbf{u}_{1}$and$\mathbf{u}_{2}$and if $W =$ Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$, then$\mathbf{z}$must be in $W^{orthogonal}$
Answer
A. The statement is true. Since$\mathbf{z}$ is orthogonal to$\mathbf{u}_{1}$ and$\mathbf{u}_{2}$, it is orthogonal to every vector in Span $\{ \mathbf{u}_{1}$,$\mathbf{u}_{2} \}$ Spanu1, u2, a set that spans W.
6.3.22
Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in $\mathbb{R}^{n}$. For each$\mathbf{y}$and each subspace W, the vector$\mathbf{y}$- pro $j_{Upper}$ W$\mathbf{y}$is orthogonal to W
Answer
B. The statement is true. Because$\mathbf{y}$ can be written uniquely in the form$\mathbf{y} y =$ pro $j_{Upper}$ W$\mathbf{y}$projWy $+ + \mathbf{z}$, where pro $j_{Upper}$ W$\mathbf{y}$projWy is in W and$\mathbf{z}$ is in $W^{orthogonal}$ W⊥, it follows that$\mathbf{z} z = \mathbf{y} y -$ - pro $j_{Upper}$ W$\mathbf{y}$projWy.
6.3.23
Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in $\mathbb{R}^{n}$. The orthogonal projection ModifyingAbove$\mathbf{y}$with caret of$\mathbf{y}$onto a subspace W can sometimes depend on the orthogonal basis for W used to compute ModifyingAbove$\mathbf{y}$with caret
Answer
B. The statement is false. The uniqueness property of the orthogonal decomposition$\mathbf{y} y =$ ModifyingAbove$\mathbf{y}$with caret $y + + \mathbf{z}$z indicates that, no matter the basis used to find it, the decomposition will always be the same.
6.3.26
Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in $\mathbb{R}^{n}$. If W is a subspace of $\mathbb{R}^{n}$ and if$\mathbf{v}$is in both W and $W^{orthogonal}$, then$\mathbf{v}$must be the zero vector
Answer
The statement is true. If$\mathbf{v}$is in W, then pro $j_{Upper}$ W$\mathbf{v}$=$\mathbf{v}$. Since the $W^{orthogonal}$ component of$\mathbf{v}$is equal to$\mathbf{v} - proj_{Upper}$ W$\mathbf{v}$, the $W^{orthogonal}$ component of$\mathbf{v}$must be$\mathbf{0}$. A similar argument can be formed for the W component of$\mathbf{v}$based on the orthogonal projection of$\mathbf{v}$onto the subspace $W^{orthogonal}$. Thus,$\mathbf{v}$must be$\mathbf{0}$
6.3.28
Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in $\mathbb{R}^{n}$. If$\mathbf{y}$=$\mathbf{z}_{1}$+$\mathbf{z}_{2}$, where$\mathbf{z}_{1}$is in a subspace W and$\mathbf{z}_{2}$is in $W^{orthogonal}$, then$\mathbf{z}_{1}$must be the orthogonal projection of$\mathbf{y}$onto W
Answer
The statement is true. Since the orthogonal decomposition of$\mathbf{y}$into components that exist in W and $W^{orthogonal}$ is unique,$\mathbf{z}_{1}$must correspond to the orthogonal projection of$\mathbf{y}$onto W