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AP PreCalculus Practice Quiz: Linear Transformations and Matrices

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

A linear transformation $L$ is represented by the matrix $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$. What is the output vector $L(\vec{v})$ for the input vector $\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$?

All Questions (10)

A linear transformation $L$ is represented by the matrix $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$. What is the output vector $L(\vec{v})$ for the input vector $\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$?

A) $\begin{pmatrix} 7 \\ 8 \end{pmatrix}$

B) $\begin{pmatrix} 6 \\ 4 \end{pmatrix}$

C) $\begin{pmatrix} 5 \\ 6 \end{pmatrix}$

D) $\begin{pmatrix} 8 \\ 7 \end{pmatrix}$

Correct Answer: A

To find the output vector, we multiply the matrix $A$ by the vector $\vec{v}$: $A\vec{v} = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3(2) + 1(1) \\ 2(2) + 4(1) \end{pmatrix} = \begin{pmatrix} 6 + 1 \\ 4 + 4 \end{pmatrix} = \begin{pmatrix} 7 \\ 8 \end{pmatrix}$. This follows the rule for determining the output vectors of a linear transformation using a $2 \\times 2$ matrix.

Which of the following functions $L$ mapping a vector $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ to an output vector is a linear transformation according to the provided definition?

A) $L(\vec{v}) = \begin{pmatrix} x^2 \\ y \end{pmatrix}$

B) $L(\vec{v}) = \begin{pmatrix} x+1 \\ y \end{pmatrix}$

C) $L(\vec{v}) = \begin{pmatrix} 5x - 2y \\ 3y \end{pmatrix}$

D) $L(\vec{v}) = \begin{pmatrix} xy \\ x \end{pmatrix}$

Correct Answer: C

A linear transformation maps an input vector to an output vector such that each component of the output vector is the sum of constant multiples of the input vector components. In option C, the first component ($5x - 2y$) and the second component ($3y$) are both sums of constant multiples of the input components $x$ and $y$. The other options include non-linear terms ($x^2$, $xy$) or the addition of a constant ($x+1$).

For a given linear transformation $L$ from $R^2$ to $R^2$, what is true about the $2 \\times 2$ matrix $A$ for which $L(\vec{v}) = A\vec{v}$?

A) The matrix $A$ is always an identity matrix.

B) There are infinitely many matrices that can represent $L$.

C) The matrix $A$ is unique for the transformation $L$.

D) The matrix $A$ only exists if the transformation is a rotation.

Correct Answer: C

The provided content states that for a linear transformation, $L$, from $R^2$ to $R^2$, there is a unique $2 \\times 2$ matrix, $A$, such that $L(\vec{v}) = A\vec{v}$. This means that for any specific linear transformation, there is only one matrix that represents it.

A linear transformation is represented by the matrix $A = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}$. This transformation is applied to a set of two input vectors, which are the columns of the matrix $V = \begin{pmatrix} 1 & 4 \\ 2 & 5 \end{pmatrix}$. What is the resulting matrix of output vectors?

A) $\begin{pmatrix} -3 & -6 \\ 6 & 15 \end{pmatrix}$

B) $\begin{pmatrix} 1 & -8 \\ 0 & 15 \end{pmatrix}$

C) $\begin{pmatrix} -3 & 6 \\ -6 & 15 \end{pmatrix}$

D) $\begin{pmatrix} 9 & 14 \\ 2 & 5 \end{pmatrix}$

Correct Answer: A

To find the matrix of output vectors, we multiply the transformation matrix $A$ by the matrix of input vectors $V$. $AV = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 1(1) + (-2)(2) & 1(4) + (-2)(5) \\ 0(1) + 3(2) & 0(4) + 3(5) \end{pmatrix} = \begin{pmatrix} 1-4 & 4-10 \\ 0+6 & 0+15 \end{pmatrix} = \begin{pmatrix} -3 & -6 \\ 6 & 15 \end{pmatrix}$.

A linear transformation $L$ maps an input vector $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ to an output vector $L(\vec{v}) = \begin{pmatrix} 4x - y \\ -x + 2y \end{pmatrix}$. What is the unique $2 \\times 2$ matrix $A$ that represents this transformation?

A) $\begin{pmatrix} 4 & 2 \\ -1 & -1 \end{pmatrix}$

B) $\begin{pmatrix} 4 & -1 \\ -1 & 2 \end{pmatrix}$

C) $\begin{pmatrix} 4 & -1 \\ 2 & -1 \end{pmatrix}$

D) $\begin{pmatrix} x & y \\ 4x-y & -x+2y \end{pmatrix}$

Correct Answer: B

We are looking for a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ such that $A\vec{v} = L(\vec{v})$. This means $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax+by \\ cx+dy \end{pmatrix} = \begin{pmatrix} 4x - y \\ -x + 2y \end{pmatrix}$. By comparing the components, we see that $a=4, b=-1, c=-1$, and $d=2$. Therefore, the matrix is $A = \begin{pmatrix} 4 & -1 \\ -1 & 2 \end{pmatrix}$.

A $2 \\times 2$ transformation matrix $A$ is multiplied by a $2 \\times 7$ matrix $V$ containing 7 input vectors. What are the dimensions of the resulting matrix of 7 output vectors?

A) $2 \\times 2$

B) $7 \\times 2$

C) $7 \\times 7$

D) $2 \\times 7$

Correct Answer: D

According to the rules of matrix multiplication and the provided content, multiplying a $2 \\times 2$ transformation matrix by a $2 \\times n$ matrix of input vectors results in a $2 \\times n$ matrix of output vectors. In this case, $n=7$, so the dimensions of the resulting matrix are $2 \\times 7$.

Let a linear transformation be represented by the matrix $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Determine the output vector when the input vector is $\vec{v} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$.

A) $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$

B) $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$

C) $\begin{pmatrix} -5 \\ -3 \end{pmatrix}$

D) $\begin{pmatrix} -3 \\ -5 \end{pmatrix}$

Correct Answer: C

The output vector is found by computing the product $A\vec{v}$. $A\vec{v} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 0(-3) + (-1)(5) \\ 1(-3) + 0(5) \end{pmatrix} = \begin{pmatrix} 0 - 5 \\ -3 + 0 \end{pmatrix} = \begin{pmatrix} -5 \\ -3 \end{pmatrix}$.

A function $L$ maps vectors in $R^2$ to $R^2$. The function is defined by a $2 \\times 2$ matrix $A$ such that $L(\vec{v}) = A\vec{v}$. Based on the provided content, why is this function guaranteed to be a linear transformation?

A) Because any function represented by a matrix is, by definition, a linear transformation.

B) Because the matrix multiplication $A\vec{v}$ results in an output where each component is a sum of constant multiples of the input vector's components.

C) Because the matrix A is unique for the function L.

D) Because the dimensions of the input and output vectors are the same.

Correct Answer: B

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$. The product $A\vec{v}$ is $\begin{pmatrix} ax+by \\ cx+dy \end{pmatrix}$. The first component, $ax+by$, is a sum of constant multiples of $x$ and $y$. The second component, $cx+dy$, is also a sum of constant multiples of $x$ and $y$. This matches the definition of a linear transformation provided in the content.

A linear transformation $L$ is applied to the standard basis vectors of $R^2$, $\vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. The transformation is represented by the matrix $A = \begin{pmatrix} 5 & -1 \\ 2 & 3 \end{pmatrix}$. What is the matrix whose columns are the output vectors $L(\vec{e}_1)$ and $L(\vec{e}_2)$?

A) $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

B) $\begin{pmatrix} 5 & 2 \\ -1 & 3 \end{pmatrix}$

C) $\begin{pmatrix} 5 & -1 \\ 2 & 3 \end{pmatrix}$

D) $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$

Correct Answer: C

We can find the output vectors by multiplying the matrix $A$ by a matrix containing the input vectors as columns: $A \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & -1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5(1)+(-1)(0) & 5(0)+(-1)(1) \\ 2(1)+3(0) & 2(0)+3(1) \end{pmatrix} = \begin{pmatrix} 5 & -1 \\ 2 & 3 \end{pmatrix}$. The resulting matrix of output vectors is the transformation matrix $A$ itself.

A linear transformation $L$ maps $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and maps $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$. What is the output of this transformation for the input vector $\vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$?

A) $\begin{pmatrix} 1 \\ 7 \end{pmatrix}$

B) $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$

C) $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$

D) $\begin{pmatrix} 4 \\ -4 \end{pmatrix}$

Correct Answer: C

The columns of the unique transformation matrix $A$ are the images of the standard basis vectors. So, $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$. To find the output for $\vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, we compute $A\vec{v}$: $\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2(2) + (-1)(-1) \\ 3(2) + 4(-1) \end{pmatrix} = \begin{pmatrix} 4+1 \\ 6-4 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$.