Chapter 3 Study Guide

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Table of Contents

Chapter 3 — Determinants

Covers §3.1–3.3. Full problems and recorded answers in 2160 Compendium.

Weighting

41 questions, only 3 true/false (7%) — the most computational chapter in the course. Points here come from executing cofactor expansion and row reduction cleanly, not from conceptual argument.

§3.1 Introduction to Determinants

Definition 1 (Cofactor expansion).

The $(i,j)$ minor $M_{ij}$ is the determinant of the matrix with row $i$ and column $j$ deleted. The cofactor is

$$C_{ij}=(-1)^{i+j}M_{ij}$$

Expansion across row $i$: $\ \det A=a_{i1}C_{i1}+a_{i2}C_{i2}+\cdots+a_{in}C_{in}$ Expansion down column $j$: $\ \det A=a_{1j}C_{1j}+a_{2j}C_{2j}+\cdots+a_{nj}C_{nj}$

Any row or column gives the same answer. That freedom is the whole strategy.

The sign pattern

$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$

Start with $+$ at the top-left and alternate. Forgetting a sign is the single most common arithmetic error in this chapter.

Choose the row or column with the most zeros

Six §3.1 problems say “at each step, choose a row or column that involves the least amount of computation.” A zero entry kills its entire cofactor term, so expanding along a row with two zeros in a $4\times4$ turns four $3\times3$ determinants into one.

Triangular shortcut

If $A$ is triangular, $\det A$ is the product of the diagonal entries. This is why elementary matrices have such easy determinants.

Tested as

Expansion across row 1 and down column 2 — 3.1.1, 3.1.3, 3.1.4, 3.1.5
Strategic expansion for minimal work — 3.1.9 – 3.1.14
Determinant of an elementary matrix — 3.1.25, 3.1.26, 3.1.29
Effect of a row operation on $\det$ — 3.1.21
Is $\det(3A)=3\det A$? — 3.1.37

§3.2 Properties of Determinants

Theorem 2 (Row operations and the determinant).

Operation	Effect on $\det$
Replacement (add a multiple of one row to another)	unchanged
Interchange two rows	multiplied by $-1$
Scale a row by $k$	multiplied by $k$

Replacement being free is what makes row reduction the efficient way to compute a large determinant.

Multiplicative properties

$$\det(AB)=(\det A)(\det B) \qquad \det(A^{T})=\det A \qquad \det(A^{-1})=\frac{1}{\det A}$$

But $\det(A+B) \neq \det A + \det B$ — the determinant is not linear in the matrix.

For an $n\times n$ matrix and scalar $r$: $\ \det(rA)=r^{n}\det A$ — the scalar comes out once per row, hence the exponent $n$. This is exactly what 3.1.37 and 3.2.42 are testing.

Invertibility

$A$ is invertible $\iff \det A \neq 0$. This is statement 12 of the Invertible Matrix Theorem.

Tested as

Name the property illustrated — 3.2.1, 3.2.2
Determinant by row reduction to echelon form — 3.2.5, 3.2.6
Row reduction combined with cofactor expansion — 3.2.12, 3.2.14
Invertibility via determinant — 3.2.21 – 3.2.23
Formula for $\det(rA)$ — 3.2.42
Verify $\det AB=(\det A)(\det B)$ — 3.2.43
Combine properties to evaluate — 3.2.45

§3.3 Cramer’s Rule, Volume, and the Adjugate

Cramer's Rule

For an invertible $n\times n$ matrix $A$, the unique solution of $A\mathbf{x}=\mathbf{b}$ has

$$x_i=\frac{\det A_i(\mathbf{b})}{\det A}$$

where $A_i(\mathbf{b})$ is $A$ with its $i$-th column replaced by $\mathbf{b}$.

Requires $\det A \neq 0$. Practical only for small systems, but it is the standard $2\times2$/$3\times3$ exam computation.

Definition 3 (Adjugate and the inverse formula).

The adjugate is the transpose of the cofactor matrix:

$$\operatorname{adj}A=\begin{bmatrix} C_{11} & C_{21} & \cdots & C_{n1} \\ C_{12} & C_{22} & \cdots & C_{n2} \\ \vdots & & & \vdots \\ C_{1n} & C_{2n} & \cdots & C_{nn} \end{bmatrix}, \qquad A^{-1}=\frac{1}{\det A}\operatorname{adj}A$$

Note the transpose — cofactor $C_{ij}$ lands in position $(j,i)$. Forgetting it is the classic error here.

Determinants as area and volume

The area of the parallelogram determined by columns of a $2\times2$ matrix $A$ is $|\det A|$
The volume of the parallelepiped determined by columns of a $3\times3$ matrix is $|\det A|$

For a parallelogram given by four vertices, first translate one vertex to the origin by subtracting it from the others, then take the two edge vectors as columns. Take the absolute value — area is never negative.

Tested as

Cramer’s Rule — 3.3.1, 3.3.2, 3.3.3, 3.3.5, 3.3.6
Adjugate and inverse formula — 3.3.12, 3.3.13, 3.3.15
Area of a parallelogram from vertices — 3.3.19, 3.3.21
Volume of a parallelepiped — 3.3.23

Graph

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