AP Precalculus Quiz: Matrices

What this quiz covers

This quiz focuses on Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $$\begin{pmatrix} 0.95 & 0.03 \\ 0.05 & 0.97 \end{pmatrix}$$ (correct answer)
2. $$\begin{pmatrix} 0.95 & 0.05 \\ 0.03 & 0.97 \end{pmatrix}$$
3. $$\begin{pmatrix} 0.05 & 0.97 \\ 0.95 & 0.03 \end{pmatrix}$$
4. $$\begin{pmatrix} -0.05 & 0.03 \\ 0.05 & -0.03 \end{pmatrix}$$

Explanation: The new urban population, $u_{k+1}$, consists of 95% of the old urban population that stays ($0.95u_k$) plus 3% of the suburban population that moves in ($0.03s_k$). So, $u_{k+1} = 0.95u_k + 0.03s_k$. The new suburban population, $s_{k+1}$, consists of 5% of the urban population that moves out ($0.05u_k$) plus 97% of the old suburban population that stays ($0.97s_k$). So, $s_{k+1} = 0.05u_k + 0.97s_k$. The transition matrix is therefore $$ \begin{pmatrix} 0.95 & 0.03 \ 0.05 & 0.97 \end{pmatrix} $$$$

Question 2

1. $\begin{pmatrix}2\\4\end{pmatrix}$ (correct answer)
2. $\begin{pmatrix}-2\\4\end{pmatrix}$
3. $\begin{pmatrix}2\\-4\end{pmatrix}$
4. $\begin{pmatrix}-4\\2\end{pmatrix}$

Explanation: This question tests AP Precalculus skills, specifically understanding matrix transformations in geometry. The given matrix A represents a 90-degree counterclockwise rotation, a fundamental transformation in computer graphics and physics. In transformation geometry, this matrix rotates vectors by swapping and negating coordinates according to the pattern (x,y) → (-y,x). Choice A is correct because Av = (0×4 + (-1)×(-2), 1×4 + 0×(-2)) = (2, 4), which represents the original vector (4,-2) rotated 90 degrees counterclockwise. Choice B incorrectly negates the first component, likely from a sign error in multiplication. To help students: Visualize transformations by plotting original and transformed vectors, and memorize common transformation matrices like rotations and reflections. Practice verifying results by checking that rotation preserves vector length.

Question 3

1. $3700$
2. $4300$ (correct answer)
3. $5300$
4. $2700$

Explanation: The initial state vector is $\vec{v}_0 = \begin{pmatrix} 3000 \\ 5000 \end{pmatrix}$. The state vector for the next month is $\vec{v}_1 = T\vec{v}_0$. We calculate: $$\vec{v}_1 = \begin{pmatrix} 0.9 & 0.2 \ 0.1 & 0.8 \end{pmatrix} \begin{pmatrix} 3000 \ 5000 \end{pmatrix} = \begin{pmatrix} 0.9(3000) + 0.2(5000) \ 0.1(3000) + 0.8(5000) \end{pmatrix} = \begin{pmatrix} 2700 + 1000 \ 300 + 4000 \end{pmatrix} = \begin{pmatrix} 3700 \ 4300 \end{pmatrix} $$$$

Question 4

1. $$\begin{pmatrix} 0 & 3 \\ 0.8 & 0.6 \end{pmatrix}$$ (correct answer)
2. $$\begin{pmatrix} 0.8 & 0 \\ 3 & 0.6 \end{pmatrix}$$
3. $$\begin{pmatrix} 0.6 & 0.8 \\ 3 & 0 \end{pmatrix}$$
4. $$\begin{pmatrix} 0 & 0.8 \\ 3 & 0.6 \end{pmatrix}$$

Explanation: The number of juveniles in the next year ($j_{k+1}$) is determined by the number of new juveniles produced by adults. Each adult produces 3 juveniles, so $j_{k+1} = 3a_k = 0j_k + 3a_k$. The number of adults in the next year ($a_{k+1}$) is determined by juveniles surviving to become adults and adults surviving. 80% of juveniles become adults ($0.8j_k$), and 60% of adults survive ($0.6a_k$). So, $a_{k+1} = 0.8j_k + 0.6a_k$. This corresponds to the matrix $$ \begin{pmatrix} 0 & 3 \ 0.8 & 0.6 \end{pmatrix} $$$$

Question 5

1. $25\%$
2. $50\%$
3. $75\%$ (correct answer)
4. $90\%$

Explanation: A steady state vector $\vec{v} = \begin{pmatrix} r \\ n \end{pmatrix}$ satisfies $T\vec{v} = \vec{v}$, where $r$ is the proportion that recycles and $n$ does not, and $r+n=1$. From $T\vec{v} = \vec{v}$, we get the equation $0.9r + 0.3n = r$. This simplifies to $0.3n = 0.1r$, or $r = 3n$. Substituting into $r+n=1$ gives $3n+n=1$, so $4n=1$, which means $n=0.25$. Then $r = 3(0.25) = 0.75$. In the steady state, 75% of households recycle.

Question 6

1. $\begin{pmatrix}14\\23\end{pmatrix}$ (correct answer)
2. $\begin{pmatrix}13\\23\end{pmatrix}$
3. $\begin{pmatrix}14\\22\end{pmatrix}$
4. $\begin{pmatrix}23\\14\end{pmatrix}$

Explanation: This question tests AP Precalculus skills, specifically understanding matrix multiplication in economic input-output models. In these models, the matrix A represents production coefficients showing how much of each sector's output is needed per unit of the other sector's production. The calculation Ad represents intermediate demand when final demand is d. Choice A is correct because Ad = (0.2×50 + 0.1×40, 0.3×50 + 0.2×40) = (10+4, 15+8) = (14, 23), showing how much of each sector's output is consumed internally. Choice B has an arithmetic error in the first component, computing 13 instead of 14. To help students: Break down matrix multiplication into dot products of rows with the vector, and interpret results in the economic context. Practice with decimal coefficients to build computational accuracy.

Question 7

1. $$\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$ (correct answer)
2. $$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
3. $$\begin{pmatrix} 17 \\ 7 \end{pmatrix}$$
4. $$\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$

Explanation: To solve for $X$, we must find the inverse of $A$ and multiply it by the resulting matrix. $X = A^{-1} \begin{pmatrix} 3 \\ 1 \end{pmatrix}$. The determinant of $A$ is $\det(A) = (5)(1) - (2)(2) = 5-4 = 1$. The inverse of $A$ is $A^{-1} = \frac{1}{1} \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$. Now, we compute $$X = \begin{pmatrix} 1 & -2 \ -2 & 5 \end{pmatrix} \begin{pmatrix} 3 \ 1 \end{pmatrix} = \begin{pmatrix} (1)(3) + (-2)(1) \ (-2)(3) + (5)(1) \end{pmatrix} = \begin{pmatrix} 3-2 \ -6+5 \end{pmatrix} = \begin{pmatrix} 1 \ -1 \end{pmatrix} $$$$

Question 8

1. $L(x, y) = (x, y)$
2. $L(x, y) = (y, x)$
3. $L(x, y) = (x, 5)$ (correct answer)
4. $L(x, y) = (0, 0)$

Explanation: A linear transformation must map the zero vector to the zero vector, i.e., $L(0,0) = (0,0)$. For choice C, $L(0,0) = (0,5)$, which is not the zero vector. Therefore, it is not a linear transformation. All other choices satisfy the properties of linear transformations. A is the identity, B is a reflection, and D is the zero transformation.

Question 9

1. $$\begin{pmatrix} 1 & 3 \\ -2 & 1 \end{pmatrix}$$
2. $$\begin{pmatrix} -5 & 3 \\ -2 & 1 \end{pmatrix}$$ (correct answer)
3. $$\begin{pmatrix} 1 & 0 \\ -2 & -5 \end{pmatrix}$$
4. $$\begin{pmatrix} 1 & 3 \\ -2 & -5 \end{pmatrix}$$

Explanation: To find the product $BA$, we multiply matrix B by matrix A. $$BA = \begin{pmatrix} 1 & 3 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ -2 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1)+(3)(-2) & (1)(0)+(3)(1) \ (0)(1)+(1)(-2) & (0)(0)+(1)(1) \end{pmatrix} = \begin{pmatrix} 1-6 & 0+3 \ 0-2 & 0+1 \end{pmatrix} = \begin{pmatrix} -5 & 3 \ -2 & 1 \end{pmatrix} $$$$

Question 10

1. $$\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$$ (correct answer)
2. $$\begin{pmatrix} -3 & 5 \\ 1 & -2 \end{pmatrix}$$
3. $$\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$$
4. $$\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}$$

Explanation: The matrix associated with the inverse transformation $L^{-1}$ is the inverse of the matrix $A$, denoted $A^{-1}$. First, we calculate the determinant of $A$: $\det(A) = (2)(3) - (5)(1) = 6-5=1$. The inverse matrix is given by $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Substituting the values, we get $$A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -5 \ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -5 \ -1 & 2 \end{pmatrix} $$$$

Question 11

1. $\det(A)=0$, so $A$ is not invertible. (correct answer)
2. $\det(A)=2$, so $A$ is not invertible.
3. $\det(A)=0$, so $A$ is invertible.
4. $\det(A)=-2$, so $A$ is invertible.

Explanation: This question tests AP Precalculus skills, specifically understanding determinants and matrix invertibility in network analysis. The determinant of a matrix indicates whether the matrix has an inverse: a matrix is invertible if and only if its determinant is non-zero. In this adjacency matrix representing network connections, we calculate det(A) by expansion: det(A) = 0×(0×0 - 0×0) - 1×(1×0 - 1×0) + 1×(1×0 - 0×1) = 0. Choice A is correct because det(A) = 0, which means the matrix is singular (not invertible), indicating the network has redundant paths or dependencies. Choice D incorrectly calculates the determinant as -2, likely from arithmetic errors in the expansion. To help students: Practice determinant calculation using cofactor expansion and verify results using row reduction. Emphasize the connection between zero determinant and linear dependence in real-world networks.

Question 12

1. $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ (correct answer)
2. $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$
3. $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
4. $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

Explanation: This question tests AP Precalculus skills, specifically identifying transformation matrices for geometric reflections. Reflection across the y-axis is a fundamental transformation that negates x-coordinates while preserving y-coordinates, transforming (x,y) to (-x,y). Choice A is correct because the matrix (-1,0; 0,1) performs exactly this transformation: when applied to (x,y), it yields (-1×x + 0×y, 0×x + 1×y) = (-x,y). Choice B represents reflection across the x-axis, while choices C and D represent different transformations entirely. To help students: Memorize standard transformation matrices and verify them by testing on simple vectors like (1,0) and (0,1). Visualize transformations by plotting several points and their images to build geometric intuition.

Question 13

1. It increases by $\begin{pmatrix}1\\3\end{pmatrix}$. (correct answer)
2. It increases by $\begin{pmatrix}3\\1\end{pmatrix}$.
3. It increases by $\begin{pmatrix}1\\1\end{pmatrix}$.
4. It increases by $\begin{pmatrix}2\\3\end{pmatrix}$.

Explanation: This question tests AP Precalculus skills, specifically understanding vector-valued functions with parameters. Vector functions can model systems where multiple quantities depend on a single parameter, common in economics and engineering. In this input-output model, x(p) gives sector outputs as functions of parameter p, where the first sector produces p+2 units and the second produces 3p-1 units. Choice A is correct because when p increases by 1, the first component changes from p+2 to (p+1)+2 = p+3, an increase of 1, and the second component changes from 3p-1 to 3(p+1)-1 = 3p+2, an increase of 3. Choice B incorrectly reverses the components, a common error when students don't carefully track which function corresponds to which sector. To help students: Emphasize systematic substitution and encourage checking work by computing specific examples. Practice interpreting parameter changes in various contexts to build intuition.

Question 14

1. $A(B+C) = AB + AC$, where $C$ is an $n \times n$ matrix.
2. $(AB)^T = B^T A^T$, where $T$ denotes the transpose.
3. $A + B = B + A$
4. $AB = BA$ (correct answer)

Explanation: Matrix multiplication is not commutative in general. While it is true for some specific matrices, it is not always true that $AB = BA$. The other three statements are standard properties of matrix operations: (A) is the distributive property, (B) is the property of the transpose of a product, and (C) is the commutative property of matrix addition.