Precalculus Quiz: Matrix Transformations Of Vectors

What this quiz covers

This quiz focuses on Matrix Transformations Of Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for 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. $\begin{pmatrix}3\\1\end{pmatrix}$
2. $\begin{pmatrix}7\\3\end{pmatrix}$ (correct answer)
3. $\begin{pmatrix}1\\9\end{pmatrix}$
4. $\begin{pmatrix}2\\4\end{pmatrix}$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. For matrix $H = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and vector $\mathbf{v} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$, we compute $Hv$ by taking the dot product of each row with $v$. First entry: $1 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7$. Second entry: $0 \cdot 1 + 1 \cdot 3 = 0 + 3 = 3$. Thus $Hv = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$. Choice B is correct because it properly computes the dot products for each row. Choice A makes an error in computing the first component, calculating $1+2 \cdot 3$ but perhaps adding wrong to get 6 or confusing. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: $a \cdot x + b \cdot y$, for row 2: $c \cdot x + d \cdot y$. To check your work, verify dimensions: a $2\times2$ matrix times a $2\times1$ vector must give a $2\times1$ vector, and each component should come from one complete dot product calculation.

Question 2

1. Matrix $L$ must be the zero matrix, so $L\vec{a} = \vec{0}$
2. The composition $L^{-1} \circ L$ equals the identity, so the result should be $\vec{a}$
3. There is an error in the problem setup; this situation is impossible (correct answer)
4. Matrix $L$ could be any invertible matrix; the given information is consistent

Explanation: If vector $\vec{a}$ is first transformed by matrix $L$, then by $L^{-1}$, the composition is $L^{-1}(L\vec{a}) = (L^{-1}L)\vec{a} = I\vec{a} = \vec{a}$. Since $L^{-1}L = I$ (the identity matrix), applying $L$ followed by $L^{-1}$ to any vector must return the original vector. Therefore, the final result should be $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$, not $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$. The given scenario is mathematically impossible. Choice A incorrectly suggests $L$ is the zero matrix (which wouldn't have an inverse). Choice B correctly states the mathematical principle but doesn't recognize the contradiction. Choice D incorrectly suggests the situation is possible.

Question 3

1. $\begin{bmatrix}3x\\x-2y\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}3x+y\\x-2y\end{bmatrix}$
3. $\begin{bmatrix}3x\\y-2x\end{bmatrix}$
4. $\begin{bmatrix}3x\\-2x+y\end{bmatrix}$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. When a matrix A multiplies a vector v (written as Av), the result is a new vector found by taking the dot product of each row of A with the column vector v: for a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and vector $\begin{bmatrix} x \\ y \end{bmatrix}$, the result is $\begin{bmatrix} ax+by \\ cx+dy \end{bmatrix}$. For arbitrary vector $\begin{bmatrix} x \\ y \end{bmatrix}$, the transformation by matrix $\begin{bmatrix} 3 & 0 \\ 1 & -2 \end{bmatrix}$ gives $\begin{bmatrix} 3x+0y \\ 1x-2y \end{bmatrix}$, showing how each component of the result depends on both components of the input through the matrix entries. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in computing the second component, using y instead of x for the first term. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: $a \cdot x + b \cdot y$, for row 2: $c \cdot x + d \cdot y$. To check your work, verify dimensions: a 2×2 matrix times a 2×1 vector must give a 2×1 vector, and each component should come from one complete dot product calculation.

Question 4

1. $\begin{bmatrix}-6\\-4\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}-1\\4\end{bmatrix}$
3. $\begin{bmatrix}6\\-4\end{bmatrix}$
4. $\begin{bmatrix}-6\\4\end{bmatrix}$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix [2 0; 0 -1] scales the x-component by 2 and reflects the y-component across the x-axis (scaling by -1), so applying it to vector [-3; 4] transforms it to [-6; -4]. Choice A is correct because it accurately performs the matrix-vector multiplication. Choice D makes an error in the second component, calculating 0·(-3) + (-1)·(-4) = 4 instead of with positive 4. Common 2×2 transformations to recognize: [1 0; 0 -1] reflects across x-axis, [0 -1; 1 0] rotates 90° counterclockwise, [k 0; 0 k] scales uniformly by k, and [1 0; 0 1] is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like [1; 0] and [0; 1]), you can determine what it does to any vector by using linearity.

Question 5

1. Reflection across the line $y = \sqrt{3}x$, followed by scaling by factor $\frac{1}{2}$
2. Reflection across the line making a 30° angle with the positive x-axis (correct answer)
3. Rotation by 120° counterclockwise, then reflection across the x-axis
4. Reflection across the x-axis, then rotation by 60° counterclockwise

Explanation: First, identify what each matrix does: $P$ reflects across the x-axis (changes $(x,y)$ to $(x,-y)$). Matrix $Q$ has the form $\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$ with $\cos\theta = \frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$, so $\theta = 60°$. This form represents reflection across a line making angle $\frac{\theta}{2} = 30°$ with the x-axis. The composition is $QP = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$. This has the form $\begin{pmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{pmatrix}$ with $\phi = -60°$, representing reflection across a line making angle $\frac{-60°}{2} = -30°$ (or equivalently 30°) with the x-axis. Choice A incorrectly describes the transformation. Choice C gives the wrong order and operations. Choice D describes $RQ$ rather than $QP$.

Question 6

1. Rotation by 60° counterclockwise; resulting vector is $\begin{pmatrix} 2 \\ 2\sqrt{3} \end{pmatrix}$ (correct answer)
2. Reflection across the line $y = \sqrt{3}x$; resulting vector is $\begin{pmatrix} 2 \\ 2\sqrt{3} \end{pmatrix}$
3. Rotation by 30° counterclockwise; resulting vector is $\begin{pmatrix} 2\sqrt{3} \\ 2 \end{pmatrix}$
4. Scaling by factor $\sqrt{3}$; resulting vector is $\begin{pmatrix} 4\sqrt{3} \\ 0 \end{pmatrix}$

Explanation: Matrix $M$ has the form of a rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ with $\theta = 60°$, so it represents a 60° counterclockwise rotation. Applying this to $\vec{u}$: $M\vec{u} = \begin{pmatrix} \cos(60°) & -\sin(60°) \\ \sin(60°) & \cos(60°) \end{pmatrix} \begin{pmatrix} 4 \\ 0 \end{pmatrix} = \begin{pmatrix} 4\cos(60°) \\ 4\sin(60°) \end{pmatrix} = \begin{pmatrix} 4 \cdot \frac{1}{2} \\ 4 \cdot \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} 2 \\ 2\sqrt{3} \end{pmatrix}$. Choice B incorrectly identifies the transformation type. Choice C has the wrong angle and swapped components. Choice D misidentifies the transformation entirely.

Question 7

1. Rotation $90^\circ$ counterclockwise about the origin
2. Reflection across the $x$-axis
3. Reflection across the line $y=x$ (correct answer)
4. Uniform scaling by a factor of 2

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ reflects vectors across the line $y=x$, so applying it to a general vector $\begin{pmatrix} x \\ y \end{pmatrix}$ transforms it to $\begin{pmatrix} y \\ x \end{pmatrix}$, which represents swapping coordinates. Choice C is correct because it accurately describes the geometric transformation. Choice A incorrectly describes the geometric transformation, claiming it rotates $90^\circ$ counterclockwise when actually it reflects across $y=x$. Common 2×2 transformations to recognize: $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ reflects across x-axis, $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ rotates $90^\circ$ counterclockwise, $\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$ scales uniformly by k, and $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$), you can determine what it does to any vector by using linearity.

Question 8

1. A reflection across the $y$-axis
2. A $90^\circ$ clockwise rotation
3. A horizontal shear: $\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x+y\\y\end{bmatrix}$ (correct answer)
4. Uniform scaling by factor $2$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix $\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$ represents a horizontal shear, transforming $\begin{bmatrix}x\\y\end{bmatrix}$ to $\begin{bmatrix}x+y\\y\end{bmatrix}$, which slides points horizontally based on their y-coordinate. Choice C is correct because it correctly identifies the geometric transformation. Choice B incorrectly describes the geometric transformation, claiming it is a 90° clockwise rotation when actually it is a shear. Common 2×2 transformations to recognize: $\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$ reflects across x-axis, $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ rotates 90° counterclockwise, $\begin{bmatrix}k & 0\\0 & k\end{bmatrix}$ scales uniformly by k, and $\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$ is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$), you can determine what it does to any vector by using linearity.

Question 9

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

Explanation: When you encounter a matrix transformation problem, you're dealing with the equation $R\vec{p} = \vec{q}$, where you need to find the original vector $\vec{p}$ that was transformed into the known result $\vec{q}$. To solve this, you need to find the inverse transformation. The matrix $R = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ represents a 90° clockwise rotation. To find $\vec{p}$, you need to apply the inverse transformation: $\vec{p} = R^{-1}\vec{q}$. For this rotation matrix, the inverse is $R^{-1} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (a 90° counterclockwise rotation). Calculating: $\vec{p} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 0(3) + (-1)(2) \\ 1(3) + 0(2) \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ This confirms answer C is correct. Let's check why the other options fail. If you multiply $R$ by choice A $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$, you get $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$, not $\vec{q}$. Choice B $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ yields $\begin{pmatrix} -3 \\ -2 \end{pmatrix}$. Choice D $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ produces $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$. Study tip: For matrix transformation problems, always verify your answer by multiplying the transformation matrix by your proposed original vector. If it doesn't equal the given result vector, you've made an error. Also, recognize that finding the original vector requires the inverse transformation.

Question 10

1. $\begin{bmatrix}8 \\-3\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}7 \\-3\end{bmatrix}$
3. $\begin{bmatrix}8 \\3\end{bmatrix}$
4. $\begin{bmatrix}5 \\2\end{bmatrix}$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. When a matrix A multiplies a vector v (written as Av), the result is a new vector found by taking the dot product of each row of A with the column vector v: for a 2×2 matrix [a b; c d] and vector [x; y], the result is [ax+by; cx+dy]. For matrix A = [2 1; -1 0] and vector v = [3; 2], we compute Av by taking the dot product of each row with v. First entry: 2·3 + 1·2 = 6 + 2 = 8. Second entry: -1·3 + 0·2 = -3 + 0 = -3. Thus Av = $\begin{bmatrix}8 \\-3\end{bmatrix}$. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in computing the first component, calculating 1·3 + 2·2 = 7 instead of 2·3 + 1·2 = 8, perhaps by transposing the matrix entries. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: a·x + b·y, for row 2: c·x + d·y. Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like [1; 0] and [0; 1]), you can determine what it does to any vector by using linearity.

Question 11

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ rotates vectors 90° counterclockwise, so applying it to vector $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ transforms it to $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$, which represents the point rotated 90°. Choice A is correct because it accurately performs the matrix-vector multiplication. Choice B makes an arithmetic error, computing something like $\begin{pmatrix} -1 \times 2 + 0 \times (-3) \\ 0 \times 2 + (-1) \times (-3) \end{pmatrix}$ but that's not accurate; it might confuse the matrix entries. Common 2×2 transformations to recognize: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ reflects across x-axis, $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ rotates 90° counterclockwise, $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ scales uniformly by k, and $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$), you can determine what it does to any vector by using linearity.

Question 12

1. $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$
2. $\begin{pmatrix} 4 \\ -5 \end{pmatrix}$
3. $\begin{pmatrix} 6 \\ -8 \end{pmatrix}$
4. $\begin{pmatrix} 8 \\ 1 \end{pmatrix}$ (correct answer)

Explanation: When you see a matrix transformation problem, remember that the columns of a matrix tell you exactly where the standard basis vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ get mapped to. Since $M$ maps $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$, the matrix must be $M = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$. The first column is where $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ goes, and the second column is where $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ goes. Now multiply $M$ by $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$: $\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 2(3) + (-1)(-2) \\ 3(3) + 4(-2) \end{pmatrix} = \begin{pmatrix} 6 + 2 \\ 9 - 8 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}$ This confirms answer D is correct. Answer A gives $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$, which you might get if you incorrectly computed the matrix multiplication. Answer B gives $\begin{pmatrix} 4 \\ -5 \end{pmatrix}$, possibly from sign errors in the calculation. Answer C gives $\begin{pmatrix} 6 \\ -8 \end{pmatrix}$, which you might get if you forgot to add the cross terms in matrix multiplication and only computed $2(3) = 6$ and $4(-2) = -8$. Key strategy: When given where standard basis vectors map, you immediately know the matrix columns. This makes finding any transformation straightforward through matrix multiplication.

Question 13

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix [0 -1; 1 0] rotates vectors 90° counterclockwise, so applying it to vector [4; -1] transforms it to [ -(-1); 4 ] = [1; 4], which represents the point rotated 90°. Choice C is correct because it accurately performs the matrix-vector multiplication. Choice D has the correct components but in reversed order, giving [4; -1] rotated clockwise instead of counterclockwise. Common 2×2 transformations to recognize: [1 0; 0 -1] reflects across x-axis, [0 -1; 1 0] rotates 90° counterclockwise, [k 0; 0 k] scales uniformly by k, and [1 0; 0 1] is the identity (no change). To check your work, verify dimensions: a 2×2 matrix times a 2×1 vector must give a 2×1 vector, and each component should come from one complete dot product calculation.

Question 14

1. $\begin{pmatrix}6\\-2\end{pmatrix}$ (correct answer)
2. $\begin{pmatrix}3\\-4\end{pmatrix}$
3. $\begin{pmatrix}5\\-1\end{pmatrix}$
4. $\begin{pmatrix}1\\-4\end{pmatrix}$

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix $\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ scales the x-component by 2 while leaving y unchanged, so applying it to vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ transforms it to $\begin{pmatrix} 6 \\ -2 \end{pmatrix}$, which represents the point scaled along the x-axis. Choice A is correct because it accurately performs the matrix-vector multiplication. Choice B makes an error in computing the first component, calculating 3 instead of 6, perhaps forgetting to multiply by 2. Common 2×2 transformations to recognize: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ reflects across x-axis, $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ rotates 90° counterclockwise, $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ scales uniformly by k, and $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$), you can determine what it does to any vector by using linearity.

Question 15

1. $S = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$ (correct answer)
2. $S = \begin{pmatrix} 4 & 2 \\ 0 & 6 \end{pmatrix}$
3. $S = \begin{pmatrix} 4 & 7 \\ 0 & 9 \end{pmatrix}$
4. $S = \begin{pmatrix} 6 & 5 \\ 0 & 9 \end{pmatrix}$

Explanation: Applying transformation $T$ twice means computing $T^2 = T \cdot T$. We have $T^2 = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2(2)+1(0) & 2(1)+1(3) \\ 0(2)+3(0) & 0(1)+3(3) \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$. Therefore $S = T^2 = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$. Choice B represents $2T$ (scalar multiplication), not $T^2$. Choice C has an error in the upper-right entry. Choice D has an error in the upper-left entry.

Question 16

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Geometric transformations like rotations, reflections, and scalings can be represented by matrices: applying the matrix to a vector transforms that vector according to the geometric operation the matrix encodes. The matrix $\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$ reflects vectors across the y-axis, so applying it to vector $\begin{pmatrix}4 \\ -2\end{pmatrix}$ transforms it to $\begin{pmatrix}-4 \\ -2\end{pmatrix}$, which represents the point reflected across the y-axis. Choice C is correct because it accurately performs the matrix-vector multiplication. Choice A returns the original vector unchanged, failing to apply the transformation at all. Common 2×2 transformations to recognize: $\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$ reflects across x-axis, $\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$ rotates 90° counterclockwise, $\begin{pmatrix}k & 0 \\ 0 & k\end{pmatrix}$ scales uniformly by k, and $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ is the identity (no change). Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like $\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\begin{pmatrix}0 \\ 1\end{pmatrix}$), you can determine what it does to any vector by using linearity.

Question 17

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. When a matrix A multiplies a vector v (written as Av), the result is a new vector found by taking the dot product of each row of A with the column vector v: for a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and vector $\begin{pmatrix} x \\ y \end{pmatrix}$, the result is $\begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$. For arbitrary vector $\begin{pmatrix} x \\ y \end{pmatrix}$, the transformation by matrix $\begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix}$ gives $\begin{pmatrix} 2x + y \\ -3x + 0y \end{pmatrix}$, showing how each component of the result depends on both components of the input through the matrix entries. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in computing the second component, calculating -3y instead of -3x using dot product. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: $a \cdot x + b \cdot y$, for row 2: $c \cdot x + d \cdot y$. To check your work, verify dimensions: a $2 \times 2$ matrix times a $2 \times 1$ vector must give a $2 \times 1$ vector, and each component should come from one complete dot product calculation.

Question 18

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. To multiply a matrix by a vector, treat the vector as a column matrix (n×1) and use standard matrix multiplication: each entry of the result vector is the dot product of the corresponding row of the matrix with the column vector. For matrix $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$ and vector $\mathbf{v} = \begin{bmatrix} 0 \\ 5 \end{bmatrix}$, we compute $A\mathbf{v}$ by taking the dot product of each row with $\mathbf{v}$. First entry: $2 \cdot 0 + (-1) \cdot 5 = 0 - 5 = -5$. Second entry: $1 \cdot 0 + 1 \cdot 5 = 0 + 5 = 5$. Thus $A\mathbf{v} = \begin{bmatrix} -5 \\ 5 \end{bmatrix}$. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in the first component, ignoring the -1 and getting positive 5. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: a·x + b·y, for row 2: c·x + d·y. To check your work, verify dimensions: a 2×2 matrix times a 2×1 vector must give a 2×1 vector, and each component should come from one complete dot product calculation.

Question 19

1. $2$ square units
2. $4$ square units (correct answer)
3. $\sqrt{3} + 1$ square units
4. $2\sqrt{3}$ square units

Explanation: The area of a figure after linear transformation is the original area multiplied by $|\det(H)|$. The determinant of $H$ is: $\det(H) = (\sqrt{3})(\sqrt{3}) - (-1)(1) = 3 - (-1) = 3 + 1 = 4$. Since the original unit square has area 1, the transformed figure has area $1 \times |4| = 4$ square units. Choice A gives half the correct answer. Choice C incorrectly adds the matrix entries rather than computing the determinant. Choice D uses $2\sqrt{3}$ which doesn't correspond to any correct calculation involving this matrix.

Question 20

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

Explanation: This question tests understanding of how matrices transform vectors through matrix-vector multiplication. To multiply a matrix by a vector, treat the vector as a column matrix (n×1) and use standard matrix multiplication: each entry of the result vector is the dot product of the corresponding row of the matrix with the column vector. For matrix A = [2 -1; 1 1] and vector v = [0; 5], we compute Av by taking the dot product of each row with v. First entry: 2·0 + (-1)·5 = 0 - 5 = -5. Second entry: 1·0 + 1·5 = 0 + 5 = 5. Thus Av = [-5; 5]. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in the first component, ignoring the -1 and getting positive 5. Key to matrix-vector multiplication: treat the vector as a column matrix and compute each entry of the result as the dot product of the corresponding row of the matrix with the vector column—for row 1: a·x + b·y, for row 2: c·x + d·y. To check your work, verify dimensions: a 2×2 matrix times a 2×1 vector must give a 2×1 vector, and each component should come from one complete dot product calculation.