Matrices
Help Questions
AP Precalculus › Matrices
In network analysis, $A=\begin{pmatrix}0&1&1\1&0&0\1&0&0\end{pmatrix}$ is an adjacency matrix; what does $\det(A)$ indicate about invertibility?
$\det(A)=-2$, so $A$ is invertible.
$\det(A)=0$, so $A$ is not invertible.
$\det(A)=0$, so $A$ is invertible.
$\det(A)=2$, so $A$ is not 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.
In an input-output model, $A=\begin{pmatrix}0.2&0.1\0.3&0.2\end{pmatrix}$ and demand $\vec d=\begin{pmatrix}50\40\end{pmatrix}$; what is $A\vec d$?
$\begin{pmatrix}23\14\end{pmatrix}$
$\begin{pmatrix}13\23\end{pmatrix}$
$\begin{pmatrix}14\22\end{pmatrix}$
$\begin{pmatrix}14\23\end{pmatrix}$
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.
In transformation geometry, which matrix corresponds to a reflection across the $y$-axis acting on vectors $\begin{pmatrix}x\y\end{pmatrix}$?
$\begin{pmatrix}0&1\1&0\end{pmatrix}$
$\begin{pmatrix}1&0\0&-1\end{pmatrix}$
$\begin{pmatrix}0&-1\1&0\end{pmatrix}$
$\begin{pmatrix}-1&0\0&1\end{pmatrix}$
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.
In transformation geometry, $A=\begin{pmatrix}0&-1\1&0\end{pmatrix}$ acts on $\vec v=\begin{pmatrix}4\-2\end{pmatrix}$; what is $A\vec v$?
$\begin{pmatrix}2\4\end{pmatrix}$
$\begin{pmatrix}2\-4\end{pmatrix}$
$\begin{pmatrix}-4\2\end{pmatrix}$
$\begin{pmatrix}-2\4\end{pmatrix}$
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.
In an input-output model, $\vec x(p)=\begin{pmatrix}p+2\3p-1\end{pmatrix}$ gives sector outputs; if $p$ increases by 1, how does $\vec x$ change?
It increases by $\begin{pmatrix}1\3\end{pmatrix}$.
It increases by $\begin{pmatrix}2\3\end{pmatrix}$.
It increases by $\begin{pmatrix}3\1\end{pmatrix}$.
It increases by $\begin{pmatrix}1\1\end{pmatrix}$.
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.