Chapter 2 Study Guide

calendar_today August 9, 2026 article 706 words access_time 5 min replay Modified August 9, 2026

Table of Contents

Chapter 2 — Matrix Algebra

Covers §2.1–2.3. Full problems and recorded answers in 2160 Compendium.

Weighting

54 questions, of which 16 (29%) are true/false. This chapter is where matrix algebra stops behaving like ordinary arithmetic, and almost every conceptual question probes one of those failures.

§2.1 Matrix Operations

Definition 1 (The operations).

Sum $A+B$ — defined only when $A$ and $B$ have the same size; entrywise. Scalar multiple $rA$ — entrywise. Product $AB$ — defined only when $\text{cols}(A)=\text{rows}(B)$. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$. Column rule: $AB=[\,A\mathbf{b}_1\ \ A\mathbf{b}_2\ \cdots\ A\mathbf{b}_p\,]$ — each column of $AB$ is $A$ times the corresponding column of $B$.

Where matrix algebra breaks

These are the source of nearly every true/false question in §2.1:

$AB \neq BA$ in general — not commutative
$AB=AC$ does not imply $B=C$ — no cancellation
$AB=0$ does not imply $A=0$ or $B=0$
$(A+B)^2 \neq A^2+2AB+B^2$ unless $A$ and $B$ commute
$(AB)^T = B^T A^T$ — the transpose reverses the order

Tested as

Compute sums/products, or explain why undefined — 2.1.1, 2.1.2, 2.1.4, 2.1.5
Size arithmetic — 2.1.7, 2.1.8
Find $k$ making $AB=BA$ — 2.1.9
Verify $AB=AC$ with $B\neq C$ — 2.1.10
Transpose identities — 2.1.23, 2.1.24
Column-relation reasoning — 2.1.26, 2.1.27
Powers $A^k$ — 2.1.47

Reasoning about columns without computing

If two columns of $B$ are equal, the corresponding columns of $AB$ are equal (by the column rule). If column 4 of $B$ is the sum of columns 1 and 5, the same relation holds in $AB$. These follow immediately from $AB=[A\mathbf{b}_1\ \cdots]$ — no arithmetic needed.

§2.2 The Inverse of a Matrix

Definition 2 (Invertibility).

$A$ ($n\times n$) is invertible if there is $A^{-1}$ with $AA^{-1}=A^{-1}A=I_n$. A non-square matrix is never invertible.

The $2\times2$ formula

For $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $\det A = ad-bc \neq 0$:

$$A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

Swap the diagonal, negate the off-diagonal, divide by the determinant. If $ad-bc=0$, $A$ is not invertible.

The row-reduction algorithm

To invert a larger matrix, row reduce $[\,A \mid I\,]$. If $A$ reduces to $I$, the right half is $A^{-1}$:

$$[\,A \mid I\,] \sim [\,I \mid A^{-1}\,]$$

If $A$ does not reduce to $I$, then $A$ is not invertible.

Order-reversing identities

$(AB)^{-1} = B^{-1}A^{-1}$ and $(A^{T})^{-1} = (A^{-1})^{T}$. A product of invertible matrices is invertible; a sum of invertible matrices need not be.

Tested as

Invert by formula — 2.2.1, 2.2.3, 2.2.4
Invert by row reduction — 2.2.39 – 2.2.42
Solve $A\mathbf{x}=\mathbf{b}$ via $A^{-1}$ — 2.2.9
Prove $ABC$ invertible — 2.2.25
Why invertibility gives independent / spanning columns — 2.2.31, 2.2.32
True/false on inverse algebra — 2.2.11 – 2.2.19

§2.3 Characterizations of Invertible Matrices

Theorem 3 (The Invertible Matrix Theorem).

For a square $n\times n$ matrix $A$, the following are equivalent — all true or all false together:

$A$ is invertible
$A$ is row equivalent to $I_n$
$A$ has $n$ pivot positions
$A\mathbf{x}=\mathbf{0}$ has only the trivial solution
The columns of $A$ are linearly independent
$\mathbf{x}\mapsto A\mathbf{x}$ is one-to-one
$A\mathbf{x}=\mathbf{b}$ has a solution for every $\mathbf{b}\in\mathbb{R}^n$
The columns of $A$ span $\mathbb{R}^n$
$\mathbf{x}\mapsto A\mathbf{x}$ is onto
There is $C$ with $CA=I$; there is $D$ with $AD=I$
$A^{T}$ is invertible
$\det A \neq 0$ (from Chapter 3)

This theorem is the spine of the whole course. It is why §1.4 spanning and §1.7 independence collapse into one condition for square matrices, and it reappears in Chapters 3, 4, 5, and 6.

The one restriction that matters

Every statement above assumes $A$ is square. For non-square matrices, independence and spanning remain genuinely different — this is the trap in most §2.3 true/false problems.

"Use as few calculations as possible"

Six §2.3 problems are phrased this way. The intent is not to row reduce. Look for a shortcut:

a zero row or column ⇒ not invertible
two identical or proportional rows/columns ⇒ not invertible
triangular matrix ⇒ invertible $\iff$ no zero on the main diagonal
$2\times2$ ⇒ just check $ad-bc$

Tested as

Invertible with minimal computation — 2.3.1, 2.3.3 – 2.3.8
IMT-based true/false — 2.3.11 – 2.3.20
Triangular matrices — 2.3.21
Inconsistency ⇒ what about $H$? — 2.3.30

Graph

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