Unit 4 · Topic 4.13

Matrices as Functions

A matrix is a function on vectors.

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Part 1 · Hook

Why this matters

A game character turns a quarter turn and then mirrors itself. Each move is a matrix, so the pair of moves should be one matrix too. It is, as long as you multiply in the right order. Thinking of matrices as functions, with inputs and outputs, tells you which order, and how to undo a move.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $f(x) = 2x$ and $g(x) = x + 1$ , what is $f(g(3))$ ?

Show the answer

Inside first: $g(3) = 4$ , then $f(4) = 8$ .

Correct: $8$ :
$7$ :
$6$ :
$9$ :

2. Which matrix rotates the plane by $\frac{\pi}{2}$ counterclockwise?

$\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$

Show the answer

Its columns are the images of $\left\langle 1, 0 \right\rangle$ and $\left\langle 0, 1 \right\rangle$ : $\left\langle 0, 1 \right\rangle$ and $\left\langle -1, 0 \right\rangle$ .

Correct: $\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$ :
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$ :
$\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$ :
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$ :

Part 4 · See it

See it first

\mathbf{v} = \left\langle 3, 1 \right\rangle

is rotated by

R

\left\langle -1, 3 \right\rangle

, then reflected by

F

\left\langle -1, -3 \right\rangle

. The single matrix for "

R

, then

F

" is

FR

. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A matrix $A$ takes an input vector $\mathbf{v}$ it returns the output $A\mathbf{v}$ , so $A$ acts as a function on vectors
Apply $B$ first, then $A$ the output is $A(B\mathbf{v}) = (AB)\mathbf{v}$ : the matrix of the composition is $AB$ , written right to left
$\det A \ne 0$ $A^{-1}$ is the matrix of the inverse function: it sends each output back to its input
$\det A = 0$ two inputs share an output, so the transformation has no inverse

Part 6 · Key ideas

Key ideas

A $2 \times 2$ matrix $A$ is a function from vectors to vectors: input $\mathbf{v}$ , output $A\mathbf{v}$ .
Composition of linear transformations: "first $B$ , then $A$ " has matrix $AB$ . The first move is on the right, next to $\mathbf{v}$ .
The inverse transformation has matrix $A^{-1}$ , and exists exactly when $\det A \ne 0$ . To find the input with output $\mathbf{w}$ , compute $A^{-1}\mathbf{w}$ .
Repeating $A$ $n$ times has matrix $A^n$ , and $\det(AB) = \det A \cdot \det B$ .

Part 7 · Misconception

A common mistake

The wrong idea: "Rotate with $R$ , then reflect with $F$ " has the matrix $RF$ , because $R$ comes first.

What actually happens: The first move acts on $\mathbf{v}$ first, so it sits next to $\mathbf{v}$ : $F(R\mathbf{v}) = (FR)\mathbf{v}$ . Read a product right to left, like $f(g(x))$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A rotation and a reflection

$R$ rotates the plane by $\frac{\pi}{2}$ counterclockwise and $F$ reflects it across the $x$ -axis.

$R = \left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right], \qquad F = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$

1. Which matrix does "rotate with $R$ , then reflect with $F$ " in one step?

$\left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$

Show the answer

Rotate first, so $R$ is on the right: $FR$ $= \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$ $= \left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$ .

Correct: $\left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$ : $FR$ : the first move on the right.
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$ : This is $RF$ : reflect first, then rotate.
$\left[ \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right]$ : This adds $R$ and $F$ ; composing multiplies.
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$ : This is $R^{-1}$ , a clockwise quarter turn.

2. Which matrix does "reflect with $F$ , then rotate with $R$ " in one step?

$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$

Show the answer

Reflect first, so $F$ is on the right: $RF$ $= \left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$ $= \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$ .

Correct: $\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$ : Reflect first, so $F$ sits on the right.
$\left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$ : This is $FR$ , the other order.
$\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$ : This is $R$ alone; the reflection is missing.
$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$ : The two moves do not cancel; $RF \ne I$ .

3. What is the image of $\left\langle 2, 1 \right\rangle$ when it is rotated with $R$ and then reflected with $F$ ?

$\left\langle -1, -2 \right\rangle$
$\left\langle -1, 2 \right\rangle$
$\left\langle 1, 2 \right\rangle$
$\left\langle 2, -1 \right\rangle$

Show the answer

$R\left[ \begin{matrix} 2 \\ 1 \end{matrix} \right] = \left[ \begin{matrix} -1 \\ 2 \end{matrix} \right]$ , then $F$ flips the second entry: $\left\langle -1, -2 \right\rangle$ . Or $FR\left[ \begin{matrix} 2 \\ 1 \end{matrix} \right]$ directly.

Correct: $\left\langle -1, -2 \right\rangle$ : Rotate, then reflect.
$\left\langle -1, 2 \right\rangle$ : This is after the rotation only.
$\left\langle 1, 2 \right\rangle$ : This is the other order, $RF$ .
$\left\langle 2, -1 \right\rangle$ : This is after the reflection only.

4. The matrix $FR = \left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right]$ is a single reflection. Across which line?

$y = -x$
$y = x$
the $y$ -axis
the $x$ -axis

Show the answer

$FR\left[ \begin{matrix} x \\ y \end{matrix} \right] = \left[ \begin{matrix} -y \\ -x \end{matrix} \right]$ : it swaps the coordinates and negates both. Points on $y = -x$ , such as $\left\langle 1, -1 \right\rangle$ , stay put.

Correct: $y = -x$ : $\left\langle 1, -1 \right\rangle \to \left\langle 1, -1 \right\rangle$ : fixed on $y = -x$ .
$y = x$ : Reflection across $y = x$ is $RF$ ; it fixes $\left\langle 1, 1 \right\rangle$ , which $FR$ sends to $\left\langle -1, -1 \right\rangle$ .
the $y$ -axis: The $y$ -axis reflection sends $\left\langle 1, 0 \right\rangle$ to $\left\langle -1, 0 \right\rangle$ ; $FR$ sends it to $\left\langle 0, -1 \right\rangle$ .
the $x$ -axis: That is $F$ alone; the rotation changes it.

5. In the product $AB\mathbf{v}$ , which transformation acts on $\mathbf{v}$ first?

$B$ , the one next to $\mathbf{v}$
$A$ , the one written first
Both at once, so the order makes no difference
Whichever has the larger determinant

Show the answer

$AB\mathbf{v} = A(B\mathbf{v})$ : $B$ acts first, like $g$ in $f(g(x))$ .

Correct: $B$ , the one next to $\mathbf{v}$ : Right to left.
$A$ , the one written first: Written first, applied last.
Both at once, so the order makes no difference: $AB$ and $BA$ usually differ, so the order matters.
Whichever has the larger determinant: Determinants do not set the order.

6. A rotation by $\frac{\pi}{6}$ counterclockwise is applied three times. Which single matrix does the same?

$\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$
$\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]$
$\left[ \begin{matrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix} \right]$

Show the answer

Three turns of $\frac{\pi}{6}$ make $\frac{\pi}{2}$ , and the rotation by $\frac{\pi}{2}$ counterclockwise has columns $(0, 1)$ and $(-1, 0)$ .

Correct: $\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]$ : $3 \cdot \frac{\pi}{6} = \frac{\pi}{2}$ .
$\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$ : This turns clockwise.
$\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]$ : This is a half turn, $\pi$ .
$\left[ \begin{matrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix} \right]$ : This is one turn of $\frac{\pi}{6}$ , not three.

Part 9 · Summary

Summary

A matrix is a function on vectors. Doing

B

and then

A

is the single matrix

AB

, read right to left, and repeating

A

gives

A^n

. When

\det A \ne 0

A^{-1}

undoes

A

and recovers an input from its output; when

\det A = 0

, two inputs share an output and nothing can undo it.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic