Linear Algebra Quiz: Geometric Matrix Transformations
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Geometric Matrix TransformationsQuestion 1 of 15

The matrix M=[2102]M = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} represents a linear transformation. When applied repeatedly, MnM^n for large nn will cause most vectors to approach which direction?

The direction of the vector (1,0)(1, 0) with exponential growth rate 2n2^n
The direction of the vector (1,1)(1, 1) with exponential growth rate 2n2^n
The direction of the vector (0,1)(0, 1) with polynomial growth rate n⋅2n−1n \cdot 2^{n-1}
All directions equally, since the eigenvalues have equal magnitude
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Linear Algebra Quiz

Linear Algebra Quiz: Geometric Matrix Transformations

Practice Geometric Matrix Transformations in Linear Algebra 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 Geometric Matrix Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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.

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Question 1

The matrix M=[2102]M = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} represents a linear transformation. When applied repeatedly, MnM^n for large nn will cause most vectors to approach which direction?

  1. The direction of the vector (1,0)(1, 0) with exponential growth rate 2n2^n (correct answer)
  2. The direction of the vector (1,1)(1, 1) with exponential growth rate 2n2^n
  3. The direction of the vector (0,1)(0, 1) with polynomial growth rate n⋅2n−1n \cdot 2^{n-1}
  4. All directions equally, since the eigenvalues have equal magnitude
Explanation: M has a repeated eigenvalue λ = 2 with geometric multiplicity 1 (only one linearly independent eigenvector (1,0)). For large n, M^n behaves like 2^n times the projection onto the eigenspace of the dominant eigenvalue. Since (1,0) is the only eigenvector direction, vectors approach this direction. Choice B uses the wrong eigenvector, Choice C confuses the eigenvector with a generalized eigenvector, and Choice D incorrectly assumes diagonalizability.

Question 2

The matrix R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} represents a counter-clockwise rotation by an angle θ\theta. What is the geometric interpretation of the transformation represented by the matrix R3R^3?

  1. A rotation by an angle of θ3\theta^3.
  2. A scaling by a factor of 3.
  3. A rotation by an angle of 3θ3\theta. (correct answer)
  4. A rotation by an angle of θ\theta applied to a vector with components cubed.
Explanation: Applying a rotation matrix RR multiple times results in a composition of rotations. Applying the rotation by θ\theta three times (R3=R⋅R⋅RR^3 = R \cdot R \cdot R) is geometrically equivalent to a single rotation by the sum of the angles, which is θ+θ+θ=3θ\theta + \theta + \theta = 3\theta. Algebraically, it is a known property that for a rotation matrix R(θ)R(\theta), [R(θ)]n=R(nθ)[R(\theta)]^n = R(n\theta). Therefore, R3R^3 represents a rotation by an angle of 3θ3\theta. Choice A is incorrect as the angles add, not multiply in this fashion. Choice B is incorrect as rotations do not scale the figure (the determinant is 1). Choice D describes a non-linear operation, but matrix multiplication represents a linear transformation.

Question 3

Consider the shear transformation SS that maps (1,0)(1,0) to (1,0)(1,0) and (0,1)(0,1) to (k,1)(k,1) for some constant k>0k > 0. If SS is applied to a region RR in the plane, and then the resulting region is reflected across the line y=xy = x, the composition of these transformations preserves which geometric property?

  1. All angles between intersecting lines within the region are preserved exactly
  2. The area of the region is preserved, but distances between points may change (correct answer)
  3. Both area and all distances between points in the region are preserved
  4. The orientation of the region is preserved, but areas are scaled by factor kk
Explanation: The shear S has matrix [[1,k],[0,1]] with determinant 1, so it preserves area. The reflection across y=x has matrix [[0,1],[1,0]] with determinant -1, so it also preserves area magnitude. The composition has determinant -1, preserving area but reversing orientation. However, neither transformation is orthogonal, so distances and angles are not preserved. Choice A is wrong because shears don't preserve angles. Choice C is wrong because distances change. Choice D has the wrong scaling factor.

Question 4

A linear transformation is represented by the matrix A=(0−220)A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}. Which of the following best describes the geometric effect of this transformation?

  1. A rotation by 90∘90^\circ counter-clockwise, followed by a reflection across the y-axis.
  2. A rotation by 90∘90^\circ counter-clockwise, followed by a uniform scaling with a factor of 2. (correct answer)
  3. A reflection across the line y=−xy=-x, followed by a uniform scaling with a factor of -2.
  4. A non-uniform scaling that doubles the y-component and negates the x-component.
Explanation: The matrix AA can be decomposed to understand its effect. We can factor out a scalar: A=(0−220)=2(0−110)A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = 2 \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. The matrix (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} is the standard matrix for a 90∘90^\circ counter-clockwise rotation about the origin. The scalar 2 represents a uniform scaling (dilation) with a factor of 2. Therefore, the transformation is a rotation by 90∘90^\circ counter-clockwise combined with a uniform scaling by a factor of 2. The order of these two operations does not matter. Choice B accurately describes this composition. The other choices describe different transformations: (A) would be (−1001)(0−110)=(0110)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. (C) involves a negative scaling factor which is not standard terminology and reflection across y=−xy=-x is (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}. (D) is not a uniform scaling.

Question 5

Consider the linear transformation TT represented by the matrix A=(3−210)A = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}. Which of the following lines is mapped to itself by this transformation (i.e., is an invariant line)?

  1. y=−xy = -x
  2. y=2xy = 2x
  3. y=xy = x (correct answer)
  4. y=−12xy = -\frac{1}{2}x
Explanation: An invariant line under a linear transformation is a line passing through the origin that is spanned by an eigenvector of the transformation matrix. We must find the eigenvectors of AA. The characteristic equation is det⁡(A−λI)=0\det(A - \lambda I) = 0, which is (3−λ)(−λ)−(−2)(1)=λ2−3λ+2=0(3-\lambda)(-\lambda) - (-2)(1) = \lambda^2 - 3\lambda + 2 = 0. Factoring gives (λ−1)(λ−2)=0(\lambda-1)(\lambda-2)=0, so the eigenvalues are λ1=1\lambda_1=1 and λ2=2\lambda_2=2. For λ1=1\lambda_1=1, we solve (A−I)v⃗=0⃗(A-I)\vec{v}=\vec{0}: (2−21−1)(xy)=(00)\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. This gives the equation x−y=0x-y=0, or y=xy=x. The corresponding eigenvector is any multiple of (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}. This corresponds to the invariant line y=xy=x. For λ2=2\lambda_2=2, we solve (A−2I)v⃗=0⃗(A-2I)\vec{v}=\vec{0}: (1−21−2)(xy)=(00)\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. This gives x−2y=0x-2y=0, or y=12xy=\frac{1}{2}x. This is another invariant line, but it is not among the options. The other choices are not spanned by eigenvectors.

Question 6

A linear transformation TT is applied to the unit circle x2+y2=1x^2 + y^2 = 1. If the matrix for TT is A=(2001/2)A = \begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}, what is the area of the resulting shape?

  1. 11
  2. π\pi (correct answer)
  3. 2π2\pi
  4. 4π4\pi
Explanation: The area of the transformed shape is the area of the original shape multiplied by the absolute value of the determinant of the transformation matrix. The original shape is the unit circle, which has an area of πr2=π(1)2=π\pi r^2 = \pi (1)^2 = \pi. The transformation matrix is A=(2001/2)A = \begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}. The determinant is det⁡(A)=(2)(1/2)−(0)(0)=1\det(A) = (2)(1/2) - (0)(0) = 1. Therefore, the area of the resulting shape (an ellipse) is the original area multiplied by ∣det⁡(A)∣|\det(A)|, which is π×∣1∣=π\pi \times |1| = \pi. The transformation stretches the circle into an ellipse with semi-axes of length 2 and 1/2, but its area remains the same as the original circle's area.

Question 7

The transformation matrix A=(10−21)A = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} is applied to a square with vertices at (0,0), (1,0), (1,1), and (0,1). Which statement accurately describes the resulting figure?

  1. It is a square with the same area.
  2. It is a rectangle that is not a square.
  3. It is a parallelogram with a larger area than the original square.
  4. It is a parallelogram with the same area as the original square. (correct answer)
Explanation: The matrix A=(10−21)A = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} represents a vertical shear. A shear transformation distorts a shape but preserves its area. The determinant of the matrix is det⁡(A)=(1)(1)−(0)(−2)=1\det(A) = (1)(1) - (0)(-2) = 1. Since the area of the resulting figure is the area of the original figure multiplied by the absolute value of the determinant, the area is preserved (1×∣1∣=11 \times |1| = 1). Let's map the vertices: (0,0)→(0,0)(0,0) \to (0,0), (1,0)→(1,−2)(1,0) \to (1,-2), (1,1)→(1,−1)(1,1) \to (1,-1), and (0,1)→(0,1)(0,1) \to (0,1). The resulting figure is a parallelogram, not a square or rectangle. Therefore, the result is a parallelogram with the same area as the original square.

Question 8

A linear transformation TT maps the vertices of the unit square [0,1]×[0,1][0,1] \times [0,1] to the vertices of a parallelogram with an area of 6. The transformation also reverses the orientation of the square (e.g., vertices listed counter-clockwise are mapped to vertices listed clockwise). Which of the following could be the matrix for TT?

  1. (3002)\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}
  2. (3264)\begin{pmatrix} 3 & 2 \\ 6 & 4 \end{pmatrix}
  3. (6001)\begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}
  4. (2−21−4)\begin{pmatrix} 2 & -2 \\ 1 & -4 \end{pmatrix} (correct answer)
Explanation: The absolute value of the determinant of a transformation matrix gives the scaling factor for area. The sign of the determinant indicates whether orientation is preserved (positive) or reversed (negative). We need a matrix AA such that det⁡(A)=−6\det(A) = -6. (A) det⁡(3002)=(3)(2)−(0)(0)=6\det\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} = (3)(2) - (0)(0) = 6. The area scaling is correct, but this preserves orientation. (B) det⁡(3264)=(3)(4)−(2)(6)=0\det\begin{pmatrix} 3 & 2 \\ 6 & 4 \end{pmatrix} = (3)(4) - (2)(6) = 0. This matrix is singular and collapses the square onto a line, resulting in an area of 0. (C) det⁡(6001)=(6)(1)−(0)(0)=6\det\begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix} = (6)(1) - (0)(0) = 6. The area scaling is correct, but this preserves orientation. (D) det⁡(2−21−4)=(2)(−4)−(−2)(1)=−8+2=−6\det\begin{pmatrix} 2 & -2 \\ 1 & -4 \end{pmatrix} = (2)(-4) - (-2)(1) = -8 + 2 = -6. This matrix scales the area by 6 and reverses the orientation, satisfying both conditions.

Question 9

A transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 is a horizontal reflection across the y-axis. Another transformation S:R2→R2S: \mathbb{R}^2 \to \mathbb{R}^2 is an orthogonal projection onto the x-axis. Which matrix represents the composite transformation of first applying TT, then applying SS (denoted S∘TS \circ T)?

  1. (−1000)\begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} (correct answer)
  2. (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
  3. (1000)\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
  4. (0000)\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
Explanation: Let's find the matrix for each transformation. The horizontal reflection across the y-axis, TT, is represented by the matrix MT=(−1001)M_T = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}. The orthogonal projection onto the x-axis, SS, zeros out the y-component of any vector, so its matrix is MS=(1000)M_S = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. The composite transformation S∘TS \circ T corresponds to the matrix product MSMTM_S M_T. MSMT=(1000)(−1001)=((1)(−1)+(0)(0)(1)(0)+(0)(1)(0)(−1)+(0)(0)(0)(0)+(0)(1))=(−1000)M_S M_T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (1)(-1)+(0)(0) & (1)(0)+(0)(1) \\ (0)(-1)+(0)(0) & (0)(0)+(0)(1) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} This transformation takes a vector (x,y)(x, y), reflects it to (−x,y)(-x, y), and then projects it to (−x,0)(-x, 0). Choice B is the matrix for TT alone. Choice C is the matrix for SS alone. Choice D is the zero matrix, which would result if the projection were onto an axis orthogonal to the result of the first transformation.

Question 10

A 2D linear transformation is defined by first reflecting a point across the y-axis, and then rotating the result 90∘90^\circ counter-clockwise about the origin. Which matrix represents this composite transformation?

  1. (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
  2. (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} (correct answer)
  3. (01−10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
  4. (−1−111)\begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix}
Explanation: Let T1T_1 be the reflection across the y-axis and T2T_2 be the 90∘90^\circ counter-clockwise rotation. The corresponding matrices are M1=(−1001)M_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} and M2=(0−110)M_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. Since the reflection is applied first, the composite transformation matrix is the product M2M1M_2 M_1. M2M1=(0−110)(−1001)=((0)(−1)+(−1)(0)(0)(0)+(−1)(1)(1)(−1)+(0)(0)(1)(0)+(0)(1))=(0−1−10)M_2 M_1 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (0)(-1)+(-1)(0) & (0)(0)+(-1)(1) \\ (1)(-1)+(0)(0) & (1)(0)+(0)(1) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} This matrix represents a reflection across the line y=−xy=-x. Choice A represents the transformations applied in the reverse order (M1M2M_1 M_2). Choice C is the matrix for a 90∘90^\circ clockwise rotation. Choice D results from incorrectly adding the matrices instead of multiplying them.

Question 11

The linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 represented by matrix [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} followed by matrix $$ \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}

  1. Reflection across the line y=xtan⁡(θ/2)y = x\tan(\theta/2)
  2. Reflection across the line y=xtan⁡(π/4−θ/2)y = x\tan(\pi/4 - \theta/2)
  3. Rotation by angle θ\theta followed by scaling by factor −1-1
  4. Reflection across the line making angle θ/2\theta/2 with the positive xx-axis (correct answer)
Explanation: The composition first rotates by θ, then reflects across the x-axis. This is equivalent to a single reflection across the line that makes angle θ/2 with the positive x-axis. Choice A uses the wrong angle formula. Choice B incorrectly applies angle subtraction. Choice C describes the operations separately rather than their geometric equivalent.

Question 12

Consider the linear transformation represented by matrix $$P = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{bmatrix}

  1. A line passing through the origin with slope 11, containing all transformed points (correct answer)
  2. The entire coordinate plane, but with all areas reduced by a factor of 14\frac{1}{4}
  3. A line segment from (−1,−1)(-1, -1) to (1,1)(1, 1) containing all transformed finite points
  4. The first quadrant only, since all matrix entries are positive
Explanation: Matrix P is a projection matrix with rank 1. P maps any vector (x,y) to ((x+y)/2, (x+y)/2) = (x+y)/2 · (1,1). This means all points are projected onto the line y = x through the origin. The image is the entire line y = x, not just a segment. Choice B is wrong because the determinant is 0, so the image has no area. Choice C incorrectly suggests a bounded segment. Choice D misunderstands that the transformation can produce negative coordinates.

Question 13

A linear transformation PP projects any vector in R2\mathbb{R}^2 orthogonally onto the line y=2xy=2x. Which of the following statements about the matrix representation of PP is false?

  1. The determinant of the matrix is 0.
  2. The matrix is symmetric (PT=PP^T = P).
  3. The trace of the matrix is 0. (correct answer)
  4. Applying the transformation twice is equivalent to applying it once (P2=PP^2 = P).
Explanation: The matrix for orthogonal projection onto the line spanned by vector a⃗\vec{a} is P=1∥a⃗∥2a⃗a⃗TP = \frac{1}{\|\vec{a}\|^2} \vec{a}\vec{a}^T. For the line y=2xy=2x, we can use a⃗=(12)\vec{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, so ∥a⃗∥2=5\|\vec{a}\|^2=5. The matrix is P=15(12)(12)=15(1224)=(1/52/52/54/5)P = \frac{1}{5} \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1/5 & 2/5 \\ 2/5 & 4/5 \end{pmatrix}. (A) det⁡(P)=(1/5)(4/5)−(2/5)(2/5)=0\det(P) = (1/5)(4/5) - (2/5)(2/5) = 0. This is true. Projections onto a line are singular. (B) The matrix is clearly symmetric. This is true for all orthogonal projections. (C) The trace is Tr(P)=1/5+4/5=1Tr(P) = 1/5 + 4/5 = 1. The statement that the trace is 0 is false. (D) P2=PP^2=P is a defining property of any projection matrix. This is true. Since we are looking for the false statement, the answer is C.

Question 14

What is the geometric interpretation of the inverse of the transformation represented by the matrix A=(1−101)A = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}?

  1. A horizontal shear that transforms the point (x,y)(x, y) to (x+y,y)(x+y, y). (correct answer)
  2. A vertical shear that transforms the point (x,y)(x, y) to (x,y−x)(x, y-x).
  3. A reflection across the line y=xy=x.
  4. The transformation is not invertible.
Explanation: The given matrix A=(1−101)A = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} represents a horizontal shear that maps a point (x,y)(x, y) to (x−y,y)(x-y, y). To find the geometric interpretation of the inverse transformation, we first need to find the inverse matrix, A−1A^{-1}. For a 2x2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the inverse is 1ad−bc(d−b−ca)\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. Here, ad−bc=(1)(1)−(−1)(0)=1ad-bc = (1)(1) - (-1)(0) = 1. So, A−1=11(1101)=(1101)A^{-1} = \frac{1}{1} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. This inverse matrix transforms a point (x,y)(x, y) to (x+y,y)(x+y, y). This is a horizontal shear in the opposite direction of the original transformation. Thus, choice A is correct. Choice B describes a vertical shear. Choice C is an incorrect transformation type. Choice D is incorrect because the determinant is 1, so the matrix is invertible.

Question 15

Which matrix transforms the square with vertices (0,0),(1,0),(1,1),(0,1)(0,0), (1,0), (1,1), (0,1) into a parallelogram where the vector representing the diagonal from the origin to (1,1)(1,1) is unchanged?

  1. (2−110)\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} (correct answer)
  2. (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
  3. (2001/2)\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}
  4. (1110)\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}
Explanation: The condition that the diagonal vector from the origin to (1,1)(1,1) is unchanged means that the vector v⃗=(11)\vec{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} is an eigenvector of the transformation matrix AA with a corresponding eigenvalue of λ=1\lambda=1. We need to check which matrix AA satisfies the condition Av⃗=1v⃗A\vec{v} = 1\vec{v}, or A(11)=(11)A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. (A) (2−110)(11)=(2−11+0)=(11)\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2-1 \\ 1+0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. This matrix satisfies the condition. (B) (100−1)(11)=(1−1)≠(11)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \neq \begin{pmatrix} 1 \\ 1 \end{pmatrix}. (C) (2001/2)(11)=(21/2)≠(11)\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1/2 \end{pmatrix} \neq \begin{pmatrix} 1 \\ 1 \end{pmatrix}. (D) (1110)(11)=(21)≠(11)\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \neq \begin{pmatrix} 1 \\ 1 \end{pmatrix}. Therefore, only the matrix in choice A leaves the vector (1,1)(1,1) unchanged.