College Board AP 2025 (style)-style practice: Section II (free response, no calculator). Let A = bmatrix 0 & -1 \\ 1 & 0 bmatrix (a 90^ rotation) and B = bmatrix 1 & 0 \\ 0 & -1 bmatrix (a reflection over the x-axis), viewed as functions on vectors. (a) Find the matrix that applies A first, then B. (b) Apply it to bmatrix 1 \\ 0 bmatrix.

A 4-point question on composing matrix functions. (a) Composition (2 points): "apply A first, then B" is the product BA (rightmost acts first): BA = bmatrix 1 & 0 \\ 0 & -1 bmatrixbmatrix 0 & -1 \\ 1 & 0 bmatrix = bmatrix 0 & -1 \\ -1 & 0 bmatrix. (b) Apply (2 points): bmatrix 0 & -1 \\ -1 & 0 bmatrixbmatrix 1 \\ 0 bmatrix = bmatrix 0 \\ -1 bmatrix. The vector is rotated to point up, then reflected to point down.

United StatesPrecalculusSyllabus dot point

How is a matrix a function that takes a vector to a vector, and what do composition and inverse mean for it?

Topic 4.13 Matrices as Functions: interpret a matrix as a function from vectors to vectors, and relate matrix multiplication to composition and the inverse matrix to the inverse function.

A focused answer to AP Precalculus Topic 4.13, covering how a matrix is a function from input vectors to output vectors, how matrix multiplication corresponds to composing these functions, and how the inverse matrix undoes the transformation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A matrix can be read as a function: feed it an input vector, and it returns an output vector by matrix-times-vector multiplication. With this lens, the operations of Unit 4 line up with the function ideas of Unit 2. Multiplying two matrices is composing the two functions, with the rightmost matrix applied first, just as $f(g(x))$ applies $g$ first. The identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the function that leaves every vector unchanged, the matrix analogue of $f(x) = x$ . The inverse matrix $A^{-1}$ is the inverse function: it reverses what $A$ does, so $A^{-1}(A\mathbf{v}) = \mathbf{v}$ and $A^{-1}A = I$ . A matrix has an inverse function exactly when its determinant is non-zero, mirroring the rule that only one-to-one functions are invertible.

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What this topic is asking
A matrix as a function on vectors
Multiplication is composition
The identity and the inverse
Why this view unifies the unit
Try this

What this topic is asking

The College Board (Topic 4.13) wants you to view a matrix as a function whose input is a vector and whose output is a vector. In this view, matrix multiplication is composition of functions, the identity matrix is the do-nothing function, and the inverse matrix is the inverse function that undoes the transformation.

A matrix as a function on vectors

So the domain and range are sets of vectors, and the matrix is the rule, exactly the function structure of Unit 2 with vectors in place of numbers.

Multiplication is composition

Composition of matrix functions is the geometric "do one transformation then another", and it is computed by multiplying the matrices in the right order.

The identity and the inverse

Composition and inverse as functions

Let $A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ (stretch $x$ by $2$ ). (a) Apply $A$ to $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ . (b) Find $A^{-1}$ and confirm it undoes $A$ .

step 1 Apply A as a function

$A\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2\cdot3 \\ 1\cdot4 \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \end{bmatrix}$ . The $x$ -component doubled; the $y$ -component stayed.

step 2 Find the inverse

$\det A = 2\cdot1 - 0 = 2$ , so $A^{-1} = \frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix}$ (halve $x$ , the undoing of stretching $x$ ).

step 3 Apply the inverse to the output

$A^{-1}\begin{bmatrix} 6 \\ 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\cdot6 \\ 1\cdot4 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ , the original vector.

step 4 Confirm the composition is the identity

$A^{-1}A = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ . The inverse function exactly reverses $A$ .

Why this view unifies the unit

A point worth stating once is that the function lens turns every matrix fact into a familiar function fact. Multiplication is composition (Topic 2.7), the identity matrix is $f(x) = x$ , the inverse matrix is $f^{-1}$ (Topic 2.8), and the invertibility condition $\det A \ne 0$ is the matrix form of one-to-one. Even non-commutativity becomes intuitive: composing functions in a different order generally gives a different function. Reading matrices as functions on vectors lets you transfer all the reasoning you built in Unit 2 to the geometry of the plane, which is the conceptual payoff of the whole matrix sequence and the basis for the modelling applications of Topic 4.14.

Try this

Q1. What does the identity matrix do to a vector? [1 point]

Cue. Nothing: $I\mathbf{v} = \mathbf{v}$ , leaving the vector unchanged.

Q2. If "apply $B$ first, then $A$ " is wanted, which product do you compute? [1 point]

Cue. $AB$ : the rightmost matrix acts first, so $B$ on the right.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part A (multiple choice, no calculator). A matrix

A

is viewed as a function on vectors. Applying

A

then

A^{-1}

to a vector

\mathbf{v}

gives which result? (A)

\mathbf{0}

(B)

\mathbf{v}

(C)

2\mathbf{v}

(D)

A\mathbf{v}

Show worked answer →

The correct answer is (B), $\mathbf{v}$ .

The inverse matrix undoes the original transformation, so $A^{-1}(A\mathbf{v}) = (A^{-1}A)\mathbf{v} = I\mathbf{v} = \mathbf{v}$ . Composing a function with its inverse returns the input unchanged, exactly as for ordinary inverse functions.

AP 2025 (style)4 marksSection II (free response, no calculator). Let

A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

90^\circ

rotation) and

B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

(a reflection over the

x

-axis), viewed as functions on vectors. (a) Find the matrix that applies

A

first, then

B

. (b) Apply it to

\begin{bmatrix} 1 \\ 0 \end{bmatrix}

Show worked answer →

A 4-point question on composing matrix functions.

(a) Composition (2 points): "apply $A$ first, then $B$ " is the product $BA$ (rightmost acts first): $BA = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$ .
(b) Apply (2 points): $\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}$ . The vector is rotated to point up, then reflected to point down.

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)