Matrix Transformations of Vectors - Precalculus

1. The matrix $R=\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ represents a $90^\circ$ counterclockwise rotation. What vector results from applying this transformation to $\mathbf{v}=\begin{pmatrix}2\\-3\end{pmatrix}$? (That is, find $R\mathbf{v}$.)
   Answer: $\begin{pmatrix}3\\2\end{pmatrix}$ 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.
2. Given matrix $A$ and vector $\mathbf{v}$, $A=\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}0\\5\end{bmatrix},$ what is $A\mathbf{v}$?
   Answer: $\begin{bmatrix}-5\\5\end{bmatrix}$ 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.
3. Consider the transformation $T(\mathbf{v})=A\mathbf{v}$ with $A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}.$ Which description best matches the geometric effect of this matrix on vectors in the plane?​
   Answer: A horizontal shear: $\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x+y\\y\end{bmatrix}$ 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 [1 1; 0 1] represents a horizontal shear, transforming [x; y] to [x + y; y], 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: [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.
4. A graphics program uses the matrix transformation $T(\mathbf{v})=A\mathbf{v}$ to move points on the screen, where $A=\begin{bmatrix}2&1\\0&-1\end{bmatrix}$ and the point is represented by the column vector $\mathbf{v}=\begin{bmatrix}3\\4\end{bmatrix}.$ What is $A\mathbf{v}$?
   Answer: $\begin{bmatrix}10\\-4\end{bmatrix}$ 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]; [0 -1]] and vector v = [3; 4], we compute Av by taking the dot product of each row with v. First entry: 2·3 + 1·4 = 6 + 4 = 10. Second entry: 0·3 + (-1)·4 = 0 - 4 = -4. Thus Av = [10; -4]. Choice A is correct because it properly computes the dot products for each row. Choice C makes an arithmetic error, computing the second component as 0·3 + 1·4 = 4 instead of with the negative sign. 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.
5. Matrix $M = \begin{pmatrix} \cos(60°) & -\sin(60°) \\ \sin(60°) & \cos(60°) \end{pmatrix}$ transforms vector $\vec{u} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}$. What type of transformation does matrix $M$ represent, and what is the resulting vector?
   Answer: Rotation by 60° counterclockwise; resulting vector is $\begin{pmatrix} 2 \\ 2\sqrt{3} \end{pmatrix}$ 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.
6. Consider the transformation $T(\mathbf{v})=A\mathbf{v}$ with $A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}.$ Which description best matches the geometric effect of this matrix on vectors in the plane?
   Answer: A horizontal shear: $\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x+y\\y\end{bmatrix}$ 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.
7. Matrix $A = \begin{pmatrix} k & 2 \\ 1 & k \end{pmatrix}$ transforms vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ to vector $\begin{pmatrix} 6 \\ 5 \end{pmatrix}$. For what value of $k$ does this transformation preserve the length of all vectors (i.e., $A$ is orthogonal)?
   Answer: There is no such value of $k$; matrix $A$ cannot be orthogonal for any real $k$ This question tests two key matrix concepts: matrix transformations and orthogonal matrices. When you see a problem asking whether a matrix "preserves length," you're dealing with orthogonal matrices, which satisfy $A^T A = I$ (where $A^T$ is the transpose and $I$ is the identity matrix). First, let's find $k$ using the transformation condition. Matrix multiplication gives us: $\begin{pmatrix} k & 2 \\ 1 & k \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2k + 2 \\ 2 + k \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$ This yields two equations: $2k + 2 = 6$ and $2 + k = 5$. Both give us $k = 2$. Now, let's check if $A$ can be orthogonal when $k = 2$: $A = \begin{pmatrix} 2 & 2 \\ 1 & 2 \end{pmatrix}$ For orthogonality, we need $A^T A = I$: $\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 2 & 2 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 6 & 8 \end{pmatrix} \neq I$ Since this doesn't equal the identity matrix, $A$ is not orthogonal when $k = 2$. Choice A incorrectly states $k = 3$ (we found $k = 2$). Choice B claims $k = 2$ makes $A$ orthogonal, but we just showed it doesn't. Choice C suggests $k = -1$ makes $A$ orthogonal, but this doesn't satisfy the transformation condition and won't yield orthogonality either. The correct answer is D: no value of $k$ makes this matrix both satisfy the transformation and be orthogonal. Study tip: Always verify both conditions separately in two-part matrix problems—satisfying one constraint doesn't guarantee the other.
8. For the transformation $T(\mathbf{v})=A\mathbf{v}$ with $A=\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix},$ where does the point $\mathbf{v}=\begin{pmatrix}4 \\-2\end{pmatrix}$ map to? (Compute $A\mathbf{v}$.)
   Answer: $\begin{pmatrix}-4\\-2\end{pmatrix}$ 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.
9. A linear transformation represented by matrix $T = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ is applied twice to vector $\vec{w} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$. If the same final result could be achieved by applying a single matrix $S$ to the original vector, what is matrix $S$?
   Answer: $S = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}$ 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.
10. Given the matrix $A=\begin{pmatrix}2 & 1\\-3 & 0\end{pmatrix}$ and a general column vector $\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix},$ what is the general form of $A\mathbf{v}$?
    Answer: $\begin{pmatrix}2x+y\\-3x\end{pmatrix}$ 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.
11. A linear transformation is defined by $T(\mathbf{v})=A\mathbf{v}$, where $A=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}0\\3\end{bmatrix}.$ What is $A\mathbf{v}$?
    Answer: $\begin{bmatrix}-3\\3\end{bmatrix}$ 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 $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and vector $\begin{bmatrix} x \\ y \end{bmatrix}$, the result is $\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}$. For matrix A = $\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$ and vector v = $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, we compute Av by taking the dot product of each row with v. First entry: 2·0 + (-1)·3 = 0 - 3 = -3. Second entry: 1·0 + 1·3 = 0 + 3 = 3. Thus Av = $\begin{bmatrix} -3 \\ 3 \end{bmatrix}$. Choice A is correct because it properly computes the dot products for each row. Choice D returns the original vector unchanged, failing to apply the transformation at all. 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.
12. A transformation $T(\mathbf{v})=A\mathbf{v}$ uses $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$ Which describes the geometric effect of this matrix on vectors in the plane?
    Answer: Reflection across the line $y=x$ 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.
13. A shear transformation is defined by $H=\begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}.$ Given $\mathbf{v}=\begin{pmatrix}1\\3\end{pmatrix},$ what is the result of the matrix-vector multiplication $H\mathbf{v}$?
    Answer: $\begin{pmatrix}7\\3\end{pmatrix}$ 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.
14. For matrix-vector multiplication, let $A=\begin{bmatrix}-1&2\\3&0\end{bmatrix} \quad \text{and} \quad \mathbf{v}=\begin{bmatrix}1\\-2\end{bmatrix}.$ What is the result of multiplying matrix $A$ by vector $\mathbf{v}$ (i.e., what is $A\mathbf{v}$)?
    Answer: $\begin{bmatrix}-5\\3\end{bmatrix}$ 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 ($n \times 1$) 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 $A = \begin{bmatrix} -1 & 2 \\ 3 & 0 \end{bmatrix}$ and vector $\mathbf{v} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$, we compute $A\mathbf{v}$ by taking the dot product of each row with $\mathbf{v}$. First entry: $(-1) \cdot 1 + 2 \cdot (-2) = -1 - 4 = -5$. Second entry: $3 \cdot 1 + 0 \cdot (-2) = 3 + 0 = 3$. Thus $A\mathbf{v} = \begin{bmatrix} -5 \\ 3 \end{bmatrix}$. Choice A is correct because it properly computes the dot products for each row. Choice B makes an error in computing the components, calculating the negatives incorrectly as $\begin{bmatrix} 5 \\ -3 \end{bmatrix}$. 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.
15. A scaling transformation is defined by $S=\begin{bmatrix}2 & 0\\0 & -1\end{bmatrix}.$ Where does the point $\mathbf{v}=\begin{bmatrix}-3\\4\end{bmatrix}$ map under this transformation (i.e., what is $S\mathbf{v}$)?
    Answer: $\begin{bmatrix}-6\\-4\end{bmatrix}$ 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 $\begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}$ scales the x-component by 2 and reflects the y-component across the x-axis (scaling by -1), so applying it to vector $\begin{bmatrix} -3 \\ 4 \end{bmatrix}$ transforms it to $\begin{bmatrix} -6 \\ -4 \end{bmatrix}$. Choice A is correct because it accurately performs the matrix-vector multiplication. Choice D makes an error in the second component, calculating $0 \cdot (-3) + (-1) \cdot (-4) = 4$ instead of with positive 4. 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.
16. Given the transformation $T(\mathbf{v})=A\mathbf{v}$ with $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix},$ what does this matrix transformation do to a general vector $\mathbf{v}=\begin{bmatrix}x\\y\end{bmatrix}$?
    Answer: $T(\mathbf{v})=\begin{bmatrix}x\\-y\end{bmatrix}$ This question tests understanding of how matrices transform vectors through matrix-vector multiplication. Matrices can be viewed as transformations: the matrix A defines a function T where T(v) = Av transforms an input vector v into an output vector Av, and this transformation is linear (preserves vector addition and scalar multiplication). For arbitrary vector $\begin{bmatrix} x \\ y \end{bmatrix}$, the transformation by matrix $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ gives $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix}$, showing how each component of the result depends on both components of the input through the matrix entries. Choice B is correct because it correctly identifies the geometric transformation. Choice A incorrectly describes the geometric transformation, claiming it reflects across the y-axis when actually it reflects across the x-axis. 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.
17. Matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ maps vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and vector $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$. What vector does $M$ map $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ to?
    Answer: $\begin{pmatrix} 8 \\ 1 \end{pmatrix}$ 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.
18. The matrix $R=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$ represents a $90^\circ$ counterclockwise rotation. If $\mathbf{v}=\begin{bmatrix}4\\-1\end{bmatrix}$, what vector results from applying the transformation $R\mathbf{v}$?
    Answer: $\begin{bmatrix}1\\4\end{bmatrix}$ 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.
19. A scaling transformation doubles the $x$-coordinate but leaves the $y$-coordinate unchanged. This is represented by $S=\begin{pmatrix}2 & 0\\0 & 1\end{pmatrix}.$ For $\mathbf{v}=\begin{pmatrix}3\\-2\end{pmatrix},$ what is $S\mathbf{v}$?
    Answer: $\begin{pmatrix}6\\-2\end{pmatrix}$ 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.
20. Given the transformation $T(\mathbf{v})=A\mathbf{v}$ with $A=\begin{bmatrix}3 & 0\\1 & -2\end{bmatrix}$ and a general column vector $\mathbf{v}=\begin{bmatrix}x\\y\end{bmatrix}$, what is the general form of $A\mathbf{v}$?
    Answer: $\begin{bmatrix}3x\\x-2y\end{bmatrix}$ 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 $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and vector $\begin{bmatrix} x \\ y \end{bmatrix}$, the result is $\begin{bmatrix} a x + b y \\ c x + d y \end{bmatrix}$. 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 $2 \times 2$ 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.

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