AP PreCalculus Practice Quiz: Matrices as Functions

Correct Answer: A

Based on the provided content, there is a direct association between a linear transformation and a matrix. Specifically, a linear transformation L can be represented as L(v) = Av, where A is the matrix associated with the transformation.

Correct Answer: B

The general matrix for a counterclockwise rotation by an angle θ is given by $\begin{pmatrix} \\cos \\theta & -\\sin \\theta \\ \\sin \\theta & \\cos \\theta \end{pmatrix}$. For a 90-degree rotation, θ = 90°. Since cos(90°) = 0 and sin(90°) = 1, the matrix becomes $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$.

Correct Answer: D

The magnitude of the dilation of regions is given by the absolute value of the determinant of the transformation matrix. The determinant of matrix A is (3)(2) - (0)(0) = 6. The new area is the original area multiplied by this dilation factor: 5 * |6| = 30.

Correct Answer: C

The content states that the matrix associated with the composition of two linear transformations is the product of the matrices. Since $L_1$ is applied first, followed by $L_2$, the corresponding matrix multiplication is $B \\times A$. The transformation is $L_2(L_1(\vec{v})) = L_2(A\vec{v}) = B(A\vec{v}) = (BA)\vec{v}$.

Correct Answer: B

According to the provided content, if a linear transformation, $L$, is given by $L(\vec{v}) = A\vec{v}$, then its inverse transformation is given by $L^{-1}(\vec{v}) = A^{-1}\vec{v}$. The inverse transformation is associated with the inverse of the matrix.

Correct Answer: A

The inverse of a linear transformation undoes the original transformation. The inverse of rotating a vector by an angle θ counterclockwise is to rotate it by θ clockwise (or by -θ counterclockwise). Therefore, the inverse of a 45-degree counterclockwise rotation is a 45-degree clockwise rotation.

Correct Answer: C

The content explicitly states that 'The absolute value of the determinant of a 2x2 transformation matrix gives the magnitude of the dilation of regions in R^2 under the transformation.' This is the scaling factor for area.

Correct Answer: A

The matrix for $T_1$ (90-degree counterclockwise rotation) is $A = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$. The composite transformation is $T_2(T_1(\vec{v}))$, which corresponds to the matrix product $BA$. So, we calculate $BA = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} (2)(0)+(0)(1) & (2)(-1)+(0)(0) \ (0)(0)+(2)(1) & (0)(-1)+(2)(0) \end{pmatrix} = \begin{pmatrix} 0 & -2 \ 2 & 0 \end{pmatrix}$.

Correct Answer: B

We are given $M\vec{w} = \begin{pmatrix} 2 \ 1 \end{pmatrix}$. To find $\vec{w}$, we must apply the inverse transformation, so $\vec{w} = M^{-1}\begin{pmatrix} 2 \ 1 \end{pmatrix}$. First, find $M^{-1}$. The determinant of M is (4)(1) - (-1)(-3) = 4 - 3 = 1. The inverse is $M^{-1} = \\frac{1}{1} \begin{pmatrix} 1 & 1 \ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 3 & 4 \end{pmatrix}$. Now, calculate $\vec{w} = \begin{pmatrix} 1 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} (1)(2)+(1)(1) \ (3)(2)+(4)(1) \end{pmatrix} = \begin{pmatrix} 3 \ 10 \end{pmatrix}$.

Correct Answer: C

We compare the given matrix to the general rotation matrix $\begin{pmatrix} \\cos \\theta & -\\sin \\theta \\ \\sin \\theta & \\cos \\theta \end{pmatrix}$. We need $\\cos \\theta = -1$ and $\\sin \\theta = 0$. This occurs when the angle θ is 180 degrees (or π radians).

Correct Answer: B

The composition of two rotations is another rotation where the angles add. A rotation of 30° followed by a rotation of 60° results in a total rotation of 30° + 60° = 90°. The resulting matrix is the one for a 90° counterclockwise rotation, $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$. This is consistent with the rule that the matrix of the composition is the product of the individual matrices.

Correct Answer: D

The scaling factor for the area is the absolute value of the determinant of the transformation matrix. The determinant of A is det(A) = (2)(2) - (4)(1) = 4 - 4 = 0. Since the determinant is 0, the area of any transformed region will be the original area multiplied by |0|, which is 0. This means the transformation collapses the 2D space onto a line.

Correct Answer: A

The original rotation matrix is $A = \begin{pmatrix} \\cos \\theta & -\\sin \\theta \\ \\sin \\theta & \\cos \\theta \end{pmatrix}$. The inverse transformation corresponds to the inverse matrix, $A^{-1}$. The determinant is $\\cos^2\\theta - (-\\sin^2\\theta) = \\cos^2\\theta + \\sin^2\\theta = 1$. The inverse is $A^{-1} = \\frac{1}{1} \begin{pmatrix} \\cos \\theta & \\sin \\theta \\ -\\sin \\theta & \\cos \\theta \end{pmatrix}$. This matrix also corresponds to a rotation by angle $-\\theta$.

Correct Answer: C

The content states that the matrix for a composition is the product of the individual matrices. Applying $L_A$ first gives $A\vec{v}$. Then applying $L_B$ to this result gives $B(A\vec{v})$. By associativity of matrix multiplication, this is equivalent to $(BA)\vec{v}$. The matrix for the composite transformation is $BA$.

Correct Answer: C

The new area is the original area multiplied by the absolute value of the determinant. So, $5 = 10 \\times |\det(A)|$. This implies $|\det(A)| = 5/10 = 0.5$. The determinant itself could be 0.5 or -0.5. Of the options provided, -0.5 is a possible value for the determinant.

Correct Answer: A

The matrix for $L_1$ (90° rotation) is $A = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$. The matrix for $L_2$ (reflection) is $B = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}$. The composition of applying $L_2$ first, then $L_1$, is represented by the matrix product $AB$. So we calculate $AB = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} = \begin{pmatrix} (0)(1)+(-1)(0) & (0)(0)+(-1)(-1) \ (1)(1)+(0)(0) & (1)(0)+(0)(-1) \end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$.