Precalculus Quiz: Matrix Transformations Of Vectors
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Matrix Transformations Of VectorsQuestion 1 of 20

A shear transformation is defined by H=(1201).H=\begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}. Given v=(13),\mathbf{v}=\begin{pmatrix}1\\3\end{pmatrix}, what is the result of the matrix-vector multiplication HvH\mathbf{v}?

(31)\begin{pmatrix}3\\1\end{pmatrix}
(73)\begin{pmatrix}7\\3\end{pmatrix}
(19)\begin{pmatrix}1\\9\end{pmatrix}
(24)\begin{pmatrix}2\\4\end{pmatrix}
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Precalculus Quiz

Precalculus Quiz: Matrix Transformations Of Vectors

Practice Matrix Transformations Of Vectors in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

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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.

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Question 1

A shear transformation is defined by H=(1201).H=\begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}. Given v=(13),\mathbf{v}=\begin{pmatrix}1\\3\end{pmatrix}, what is the result of the matrix-vector multiplication HvH\mathbf{v}?

  1. (31)\begin{pmatrix}3\\1\end{pmatrix}
  2. (73)\begin{pmatrix}7\\3\end{pmatrix} (correct answer)
  3. (19)\begin{pmatrix}1\\9\end{pmatrix}
  4. (24)\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=(1201)H = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} and vector v=(13)\mathbf{v} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, we compute HvHv by taking the dot product of each row with vv. First entry: 1⋅1+2⋅3=1+6=71 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7. Second entry: 0⋅1+1⋅3=0+3=30 \cdot 1 + 1 \cdot 3 = 0 + 3 = 3. Thus Hv=(73)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⋅31+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⋅x+b⋅ya \cdot x + b \cdot y, for row 2: c⋅x+d⋅yc \cdot x + d \cdot y. To check your work, verify dimensions: a 2×22\times2 matrix times a 2×12\times1 vector must give a 2×12\times1 vector, and each component should come from one complete dot product calculation.

Question 2

Vector a⃗=(31)\vec{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} is transformed by matrix LL, and the result is then transformed by matrix L−1L^{-1}. If the final result is $$ \begin{pmatrix} 5 \ -2 \end{pmatrix}

  1. Matrix LL must be the zero matrix, so La⃗=0⃗L\vec{a} = \vec{0}
  2. The composition L−1∘LL^{-1} \circ L equals the identity, so the result should be a⃗\vec{a}
  3. There is an error in the problem setup; this situation is impossible (correct answer)
  4. Matrix LL could be any invertible matrix; the given information is consistent
Explanation: If vector a⃗\vec{a} is first transformed by matrix LL, then by L−1L^{-1}, the composition is L−1(La⃗)=(L−1L)a⃗=Ia⃗=a⃗L^{-1}(L\vec{a}) = (L^{-1}L)\vec{a} = I\vec{a} = \vec{a}. Since L−1L=IL^{-1}L = I (the identity matrix), applying LL followed by L−1L^{-1} to any vector must return the original vector. Therefore, the final result should be (31)\begin{pmatrix} 3 \\ 1 \end{pmatrix}, not (5−2)\begin{pmatrix} 5 \\ -2 \end{pmatrix}. The given scenario is mathematically impossible. Choice A incorrectly suggests LL 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

Given the transformation T(v)=AvT(\mathbf{v})=A\mathbf{v} with A=[301−2]A=\begin{bmatrix}3 & 0\\1 & -2\end{bmatrix} and a general column vector v=[xy]\mathbf{v}=\begin{bmatrix}x\\y\end{bmatrix}, what is the general form of AvA\mathbf{v}?

  1. [3xx−2y]\begin{bmatrix}3x\\x-2y\end{bmatrix} (correct answer)
  2. [3x+yx−2y]\begin{bmatrix}3x+y\\x-2y\end{bmatrix}
  3. [3xy−2x]\begin{bmatrix}3x\\y-2x\end{bmatrix}
  4. [3x−2x+y]\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 [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} and vector [xy]\begin{bmatrix} x \\ y \end{bmatrix}, the result is [ax+bycx+dy]\begin{bmatrix} ax+by \\ cx+dy \end{bmatrix}. For arbitrary vector [xy]\begin{bmatrix} x \\ y \end{bmatrix}, the transformation by matrix [301−2]\begin{bmatrix} 3 & 0 \\ 1 & -2 \end{bmatrix} gives [3x+0y1x−2y]\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⋅x+b⋅ya \cdot x + b \cdot y, for row 2: c⋅x+d⋅yc \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

A scaling transformation is defined by S=[200−1].S=\begin{bmatrix}2 & 0\\0 & -1\end{bmatrix}. Where does the point v=[−34]\mathbf{v}=\begin{bmatrix}-3\\4\end{bmatrix} map under this transformation (i.e., what is SvS\mathbf{v})?

  1. [−6−4]\begin{bmatrix}-6\\-4\end{bmatrix} (correct answer)
  2. [−14]\begin{bmatrix}-1\\4\end{bmatrix}
  3. [6−4]\begin{bmatrix}6\\-4\end{bmatrix}
  4. [−64]\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

A vector v⃗\vec{v} is transformed by matrix P=(100−1)P = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , then by matrix $$Q = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}

  1. Reflection across the line y=3xy = \sqrt{3}x, followed by scaling by factor 12\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: PP reflects across the x-axis (changes (x,y)(x,y) to (x,−y)(x,-y)). Matrix QQ has the form (cos⁡θsin⁡θsin⁡θ−cos⁡θ)\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} with cos⁡θ=12\cos\theta = \frac{1}{2} and sin⁡θ=32\sin\theta = \frac{\sqrt{3}}{2}, so θ=60°\theta = 60°. This form represents reflection across a line making angle θ2=30°\frac{\theta}{2} = 30° with the x-axis. The composition is QP=(123232−12)(100−1)=(12−323212)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 (cos⁡ϕsin⁡ϕsin⁡ϕ−cos⁡ϕ)\begin{pmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{pmatrix} with ϕ=−60°\phi = -60°, representing reflection across a line making angle −60°2=−30°\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 RQRQ rather than QPQP.

Question 6

Matrix M=(cos⁡(60°)−sin⁡(60°)sin⁡(60°)cos⁡(60°))M = \begin{pmatrix} \cos(60°) & -\sin(60°) \\ \sin(60°) & \cos(60°) \end{pmatrix} transforms vector u⃗=(40)\vec{u} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} . What type of transformation does matrix MM represent, and what is the resulting vector?

  1. Rotation by 60° counterclockwise; resulting vector is $$ \begin{pmatrix} 2 \ 2\sqrt{3} \end{pmatrix} (correct answer)
  2. Reflection across the line y=3xy = \sqrt{3}x; resulting vector is $$ \begin{pmatrix} 2 \ 2\sqrt{3} \end{pmatrix}
  3. Rotation by 30° counterclockwise; resulting vector is $$ \begin{pmatrix} 2\sqrt{3} \ 2 \end{pmatrix}
  4. Scaling by factor 3\sqrt{3}; resulting vector is $$ \begin{pmatrix} 4\sqrt{3} \ 0 \end{pmatrix}
Explanation: Matrix MM has the form of a rotation matrix (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} with θ=60°\theta = 60°, so it represents a 60° counterclockwise rotation. Applying this to u⃗\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}

Question 7

Vector p⃗=(xy)\vec{p} = \begin{pmatrix} x \\ y \end{pmatrix} undergoes transformation by matrix R=(01−10)R = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} to produce q⃗=(32)\vec{q} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} . What was the original vector p⃗\vec{p}?

  3. (correct answer)
Explanation: When you encounter a matrix transformation problem, you're dealing with the equation Rp⃗=q⃗R\vec{p} = \vec{q}, where you need to find the original vector p⃗\vec{p} that was transformed into the known result q⃗\vec{q}. To solve this, you need to find the inverse transformation. The matrix R=(01−10)R = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} represents a 90° clockwise rotation. To find p⃗\vec{p}, you need to apply the inverse transformation: p⃗=R−1q⃗\vec{p} = R^{-1}\vec{q}. For this rotation matrix, the inverse is $$R^{-1} = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} Calculating: $$\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 RR by choice A (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix}, you get (−2−3)\begin{pmatrix} -2 \\ -3 \end{pmatrix}, not q⃗\vec{q}. Choice B (2−3)\begin{pmatrix} 2 \\ -3 \end{pmatrix} yields (−3−2)\begin{pmatrix} -3 \\ -2 \end{pmatrix}. Choice D (−32)\begin{pmatrix} -3 \\ 2 \end{pmatrix} produces $$ \begin{pmatrix} 2 \ 3 \end{pmatrix} ∗∗Studytip:∗∗Formatrixtransformationproblems,alwaysverifyyouranswerbymultiplyingthetransformationmatrixbyyourproposedoriginalvector.Ifitdoesn′tequalthegivenresultvector,you′vemadeanerror.Also,recognizethatfindingtheoriginalvectorrequirestheinversetransformation. **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 8

A graphics program applies the linear transformation T(v)=AvT(\mathbf{v})=A\mathbf{v} to points in the plane, where A=[21−10],v=[32].A=\begin{bmatrix}2 & 1 \\-1 & 0\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}3 \\2\end{bmatrix}. What is AvA\mathbf{v}?

  1. [8−3]\begin{bmatrix}8 \\-3\end{bmatrix} (correct answer)
  2. [7−3]\begin{bmatrix}7 \\-3\end{bmatrix}
  3. [83]\begin{bmatrix}8 \\3\end{bmatrix}
  4. [52]\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 = [8−3]\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 9

The matrix R=(0−110)R=\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix} represents a 90∘90^\circ counterclockwise rotation. What vector results from applying this transformation to v=(2−3)\mathbf{v}=\begin{pmatrix}2\\-3\end{pmatrix} ? (That is, find RvR\mathbf{v}.)

  1. (32)\begin{pmatrix}3\\2\end{pmatrix} (correct answer)
  2. (−2−3)\begin{pmatrix}-2\\-3\end{pmatrix}
  3. (−32)\begin{pmatrix}-3\\2\end{pmatrix}
  4. (23)\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 (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} rotates vectors 90° counterclockwise, so applying it to vector (2−3)\begin{pmatrix} 2 \\ -3 \end{pmatrix} transforms it to (32)\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 (−1×2+0×(−3)0×2+(−1)×(−3))\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: (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} reflects across x-axis, (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} rotates 90° counterclockwise, (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} scales uniformly by k, and (1001)\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 (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}), you can determine what it does to any vector by using linearity.

Question 10

Matrix M=(abcd)M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} maps vector (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} to (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix} and vector (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix} to (−14)\begin{pmatrix} -1 \\ 4 \end{pmatrix}. What vector does MM map $$ \begin{pmatrix} 3 \ -2 \end{pmatrix}

  1. (5−2)\begin{pmatrix} 5 \\ -2 \end{pmatrix}
  2. (4−5)\begin{pmatrix} 4 \\ -5 \end{pmatrix}
  3. (6−8)\begin{pmatrix} 6 \\ -8 \end{pmatrix}
  4. (81)\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 (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and $$ \begin{pmatrix} 0 \ 1 \end{pmatrix} Since $$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 11

The matrix

represents a 90∘90^\circ counterclockwise rotation. If v=[4−1]\mathbf{v}=\begin{bmatrix}4\\-1\end{bmatrix}, what vector results from applying the transformation RvR\mathbf{v}?

  1. [−41]\begin{bmatrix}-4\\1\end{bmatrix}
  2. [−14]\begin{bmatrix}-1\\4\end{bmatrix}
  3. [14]\begin{bmatrix}1\\4\end{bmatrix} (correct answer)
  4. [4−1]\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 12

A scaling transformation doubles the xx-coordinate but leaves the yy-coordinate unchanged. This is represented by S=(2001).S=\begin{pmatrix}2 & 0\\0 & 1\end{pmatrix}. For v=(3−2),\mathbf{v}=\begin{pmatrix}3\\-2\end{pmatrix}, what is SvS\mathbf{v}?

  1. (6−2)\begin{pmatrix}6\\-2\end{pmatrix} (correct answer)
  2. (3−4)\begin{pmatrix}3\\-4\end{pmatrix}
  3. (5−1)\begin{pmatrix}5\\-1\end{pmatrix}
  4. (1−4)\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 (2001)\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} scales the x-component by 2 while leaving y unchanged, so applying it to vector (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix} transforms it to (6−2)\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: (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} reflects across x-axis, (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} rotates 90° counterclockwise, (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} scales uniformly by k, and (1001)\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 (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}), you can determine what it does to any vector by using linearity.

Question 13

A linear transformation represented by matrix T=(2103)T = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} is applied twice to vector w⃗=(1−2)\vec{w} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} . If the same final result could be achieved by applying a single matrix SS to the original vector, what is matrix SS?

  1. (correct answer)
Explanation: Applying transformation TT twice means computing T2=T⋅TT^2 = T \cdot T. We have T2=(2103)(2103)=(2(2)+1(0)2(1)+1(3)0(2)+3(0)0(1)+3(3))=(4509)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=T2=(4509)S = T^2 = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix} . Choice B represents 2T2T (scalar multiplication), not T2T^2. Choice C has an error in the upper-right entry. Choice D has an error in the upper-left entry.

Question 14

For the transformation T(v)=AvT(\mathbf{v})=A\mathbf{v} with A=(−1001),A=\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}, where does the point v=(4−2)\mathbf{v}=\begin{pmatrix}4 \\-2\end{pmatrix} map to? (Compute AvA\mathbf{v}.)

  1. (42)\begin{pmatrix}4\\2\end{pmatrix}
  2. (−42)\begin{pmatrix}-4\\2\end{pmatrix}
  3. (−4−2)\begin{pmatrix}-4\\-2\end{pmatrix} (correct answer)
  4. (2−4)\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 (−1001)\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} reflects vectors across the y-axis, so applying it to vector (4−2)\begin{pmatrix}4 \\ -2\end{pmatrix} transforms it to (−4−2)\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: (100−1)\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} reflects across x-axis, (0−110)\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} rotates 90° counterclockwise, (k00k)\begin{pmatrix}k & 0 \\ 0 & k\end{pmatrix} scales uniformly by k, and (1001)\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 (10)\begin{pmatrix}1 \\ 0\end{pmatrix} and (01)\begin{pmatrix}0 \\ 1\end{pmatrix}), you can determine what it does to any vector by using linearity.

Question 15

Matrix H=(3−113)H = \begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix} transforms the unit square with vertices at (0,0)(0,0), (1,0)(1,0), (1,1)(1,1), and (0,1)(0,1). What is the area of the transformed figure?

  1. 22 square units
  2. 44 square units (correct answer)
  3. 3+1\sqrt{3} + 1 square units
  4. 232\sqrt{3} square units
Explanation: The area of a figure after linear transformation is the original area multiplied by ∣det⁡(H)∣|\det(H)|. The determinant of HH is: det⁡(H)=(3)(3)−(−1)(1)=3−(−1)=3+1=4\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×∣4∣=41 \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 232\sqrt{3} which doesn't correspond to any correct calculation involving this matrix.

Question 16

Given matrix AA and vector v\mathbf{v}, A=[2−111],v=[05],A=\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}0\\5\end{bmatrix}, what is AvA\mathbf{v}?​

  1. [−55]\begin{bmatrix}-5\\5\end{bmatrix} (correct answer)
  2. [55]\begin{bmatrix}5\\5\end{bmatrix}
  3. [−50]\begin{bmatrix}-5\\0\end{bmatrix}
  4. [05]\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.

Question 17

For matrix AA and vector v\mathbf{v}, A=(30−12),v=(14),A=\begin{pmatrix}3 & 0\\-1 & 2\end{pmatrix},\quad \mathbf{v}=\begin{pmatrix}1\\4\end{pmatrix}, what is the result of multiplying AA by v\mathbf{v} (that is, what is AvA\mathbf{v})?

  1. (37)\begin{pmatrix}3\\7\end{pmatrix} (correct answer)
  2. (127)\begin{pmatrix}12\\7\end{pmatrix}
  3. (3−7)\begin{pmatrix}3\\-7\end{pmatrix}
  4. (73)\begin{pmatrix}7\\3\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 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} and vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}, the result is (ax+bycx+dy)\begin{pmatrix} a x + b y \\ c x + d y \end{pmatrix}. For matrix A = (30−12)\begin{pmatrix} 3 & 0 \\ -1 & 2 \end{pmatrix} and vector v = (14)\begin{pmatrix} 1 \\ 4 \end{pmatrix}, we compute Av by taking the dot product of each row with v. First entry: 3⋅1+0⋅4=3+0=33 \cdot 1 + 0 \cdot 4 = 3 + 0 = 3. Second entry: (−1)⋅1+2⋅4=−1+8=7(-1) \cdot 1 + 2 \cdot 4 = -1 + 8 = 7. Thus Av = (37)\begin{pmatrix} 3 \\ 7 \end{pmatrix}. 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 3∗43*4 or something to get 12 instead of 3. 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⋅ya \cdot x + b \cdot y, for row 2: c⋅x+d⋅yc \cdot x + d \cdot y. To check your work, verify dimensions: a 2×22 \times 2 matrix times a 2×12 \times 1 vector must give a 2×12 \times 1 vector, and each component should come from one complete dot product calculation.

Question 18

Consider the transformation T(v)=AvT(\mathbf{v})=A\mathbf{v} with A=[1101].A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}. Which description best matches the geometric effect of this matrix on vectors in the plane?​

  1. A reflection across the yy-axis
  2. A 90∘90^\circ clockwise rotation
  3. A horizontal shear: [xy]↦[x+yy]\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x+y\\y\end{bmatrix} (correct answer)
  4. Uniform scaling by factor 22
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 [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.

Question 19

The matrix R=[0−110]R=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix} represents a 90∘90^\circ counterclockwise rotation. If v=[4−1]\mathbf{v}=\begin{bmatrix}4\\-1\end{bmatrix}, what vector results from applying the transformation RvR\mathbf{v}?

  1. [−41]\begin{bmatrix}-4\\1\end{bmatrix}
  2. [−14]\begin{bmatrix}-1\\4\end{bmatrix}
  3. [14]\begin{bmatrix}1\\4\end{bmatrix} (correct answer)
  4. [4−1]\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 [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} rotates vectors 90° counterclockwise, so applying it to vector [4−1]\begin{bmatrix} 4 \\ -1 \end{bmatrix} transforms it to [−(−1)4]=[14]\begin{bmatrix} -(-1) \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}, 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]\begin{bmatrix} 4 \\ -1 \end{bmatrix} rotated clockwise instead of counterclockwise. Common 2×2 transformations to recognize: [100−1]\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} reflects across x-axis, [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} rotates 90° counterclockwise, [k00k]\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} scales uniformly by k, and [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} is the identity (no change). To check your work, verify dimensions: a 2×22 \times 2 matrix times a 2×12 \times 1 vector must give a 2×12 \times 1 vector, and each component should come from one complete dot product calculation.

Question 20

A graphics program applies the linear transformation T(v)=AvT(\mathbf{v})=A\mathbf{v} to points in the plane, where A=[21−10],v=[32].A=\begin{bmatrix}2 & 1\\-1 & 0\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}3\\2\end{bmatrix}. What is AvA\mathbf{v}?

  1. [8−3]\begin{bmatrix}8\\-3\end{bmatrix} (correct answer)
  2. [7−3]\begin{bmatrix}7\\-3\end{bmatrix}
  3. [83]\begin{bmatrix}8\\3\end{bmatrix}
  4. [52]\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 [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} and vector [xy]\begin{bmatrix} x \\ y \end{bmatrix}, the result is [ax+bycx+dy]\begin{bmatrix} a x + b y \\ c x + d y \end{bmatrix}. For matrix A = [21−10]\begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix} and vector v = [32]\begin{bmatrix} 3 \\ 2 \end{bmatrix}, we compute Av by taking the dot product of each row with v. First entry: 2⋅3+1⋅2=6+2=82 \cdot 3 + 1 \cdot 2 = 6 + 2 = 8. Second entry: −1⋅3+0⋅2=−3+0=−3-1 \cdot 3 + 0 \cdot 2 = -3 + 0 = -3. Thus Av = [8−3]\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=71 \cdot 3 + 2 \cdot 2 = 7 instead of 2⋅3+1⋅2=82 \cdot 3 + 1 \cdot 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⋅ya \cdot x + b \cdot y, for row 2: c⋅x+d⋅yc \cdot x + d \cdot y. Remember that matrix transformations are linear: if you understand what the matrix does to the basis vectors (like [10]\begin{bmatrix} 1 \\ 0 \end{bmatrix} and [01]\begin{bmatrix} 0 \\ 1 \end{bmatrix}), you can determine what it does to any vector by using linearity.