Geometry Quiz: Using 2x2 Matrices For Plane Transformations
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Using 2x2 Matrices For Plane TransformationsQuestion 1 of 20

A shear transformation is given by H=(1201)H=\begin{pmatrix}1&2\\0&1\end{pmatrix} . What is the image of the point (1,3)(1,3) under HH?

(7,3)(7,3)
(3,7)(3,7)
(1,5)(1,5)
(5,1)(5,1)
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Geometry Quiz: Using 2x2 Matrices For Plane Transformations

Practice Using 2x2 Matrices For Plane Transformations in Geometry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Using 2x2 Matrices For Plane Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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Question 1

A shear transformation is given by H=(1201)H=\begin{pmatrix}1&2\\0&1\end{pmatrix} . What is the image of the point (1,3)(1,3) under HH?

  1. (7,3)(7,3) (correct answer)
  2. (3,7)(3,7)
  3. (1,5)(1,5)
  4. (5,1)(5,1)
Explanation: This question tests your ability to use 2×22 \times 2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×22 \times 2 matrix can represent a linear transformation of the plane: to transform a point (x,y)(x, y), write it as column vector [x;y][x; y] and multiply by transformation matrix T=(abcd)T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} using matrix multiplication: T[x;y]=(abcd)[x;y]=[ax+by;cx+dy]=[x′;y′]T[x; y] = \begin{pmatrix} a & b \\ c & d \end{pmatrix} [x; y] = [a x + b y; c x + d y] = [x'; y'] where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ\theta counterclockwise = (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} (example: 90° rotation uses θ=90°\theta=90° giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} since cos⁡(90°)=0\cos(90°)=0 and sin⁡(90°)=1\sin(90°)=1), REFLECTION across x-axis = (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (keeps x same, negates y), REFLECTION across y-axis = (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (negates x, keeps y same), SCALING by factor k = (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! The shear matrix H=(1201)H = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} applied to (1,3)(1, 3) gives x′=1⋅1+2⋅3=7x' = 1 \cdot 1 + 2 \cdot 3 = 7, y′=0⋅1+1⋅3=3y' = 0 \cdot 1 + 1 \cdot 3 = 3, shifting x based on y while keeping y fixed, characteristic of a horizontal shear. Choice A correctly calculates this, adding twice the y-value to x for the shear effect. Distractors like choice B might swap coordinates or ignore the off-diagonal term, but remember off-diagonals introduce dependence between x and y in transformations like shears. Matrix multiplication for transformations: Given transformation matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} and point (x,y)(x, y): (1) Write point as column vector [x;y][x; y]. (2) Multiply: first row of matrix times vector gives x'-coordinate = a⋅x+b⋅ya \cdot x + b \cdot y. Second row times vector gives y'-coordinate = c⋅x+d⋅yc \cdot x + d \cdot y. (3) Result is transformed point (x′,y′)=(ax+by,cx+dy)(x', y') = (a x + b y, c x + d y). Example: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} applied to (5,3)(5, 3): x′=0⋅5+(−1)⋅3=−3x' = 0 \cdot 5 + (-1) \cdot 3 = -3, y′=1⋅5+0⋅3=5y' = 1 \cdot 5 + 0 \cdot 3 = 5, so image is (−3,5)(-3, 5). That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} is 90° rotation, (−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} is 180° rotation. REFLECTION matrices have form (±100±1)\begin{pmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{pmatrix} with exactly one negative—(100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} reflects across x-axis (y negated), (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} reflects across y-axis (x negated). SCALING matrices have equal diagonal entries (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}—both coordinates multiplied by same k, or different entries (a00b)\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). Impressive work with shears—visualize how shapes distort to understand better!

Question 2

Which matrix represents a rotation of 180∘180^\circ about the origin?

  1. (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}
  2. (−100−1)\begin{pmatrix}-1&0\\0&-1\end{pmatrix} (correct answer)
  3. (100−1)\begin{pmatrix}1&0\\0&-1\end{pmatrix}
  4. (−1001)\begin{pmatrix}-1&0\\0&1\end{pmatrix}
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = [cos(θ) -sin(θ); sin(θ) cos(θ)] (example: 90° rotation uses θ=90° giving [0 -1; 1 0] since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = [1 0; 0 -1] (keeps x same, negates y), REFLECTION across y-axis = [-1 0; 0 1] (negates x, keeps y same), SCALING by factor k = [k 0; 0 k] (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! A 180° rotation matrix is [-1 0; 0 -1], as it matches [cos(180°) -sin(180°); sin(180°) cos(180°)] = [-1 0; 0 -1], negating both coordinates to flip the point through the origin. Choice B correctly identifies this matrix by recognizing the equal negative diagonals and zero off-diagonals, distinct from reflections or 90° rotations. Distractors like choice A might confuse with 90° rotation, but note the off-diagonal signs and values differ—use the general rotation formula to verify angles. Matrix multiplication for transformations: Given transformation matrix [a b; c d] and point (x, y): (1) Write point as column vector [x; y]. (2) Multiply: first row of matrix times vector gives x'-coordinate = a·x + b·y. Second row times vector gives y'-coordinate = c·x + d·y. (3) Result is transformed point (x', y') = (ax+by, cx+dy). Example: [0 -1; 1 0] applied to (5, 3): x' = 0·5 + (-1)·3 = -3, y' = 1·5 + 0·3 = 5, so image is (-3, 5). That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form [cos(θ) -sin(θ); sin(θ) cos(θ)]—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: [0 -1; 1 0] is 90° rotation, [-1 0; 0 -1] is 180° rotation. REFLECTION matrices have form [±1 0; 0 ±1] with exactly one negative—[1 0; 0 -1] reflects across x-axis (y negated), [-1 0; 0 1] reflects across y-axis (x negated). SCALING matrices have equal diagonal entries [k 0; 0 k]—both coordinates multiplied by same k, or different entries [a 0; 0 b] for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). Wonderful progress on rotations—experiment with different angles!

Question 3

Which matrix represents a 180∘180^\circ rotation about the origin?

  1. [0−110]\begin{bmatrix}0&-1\\1&0\end{bmatrix}
  2. [−100−1]\begin{bmatrix}-1&0\\0&-1\end{bmatrix} (correct answer)
  3. [100−1]\begin{bmatrix}1&0\\0&-1\end{bmatrix}
  4. [−1001]\begin{bmatrix}-1&0\\0&1\end{bmatrix}
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = [cos(θ) -sin(θ); sin(θ) cos(θ)] (example: 90° rotation uses θ=90° giving [0 -1; 1 0] since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = [1 0; 0 -1] (keeps x same, negates y), REFLECTION across y-axis = [-1 0; 0 1] (negates x, keeps y same), SCALING by factor k = [k 0; 0 k] (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! A 180° rotation transforms (x, y) to (-x, -y), matching the matrix [-1 0; 0 -1] from the rotation formula with θ=180°. Choice B correctly identifies this matrix by recognizing the pattern of -1 on both diagonals. A distractor like choice A might confuse it with 90° rotation, which has off-diagonal entries. Matrix multiplication for transformations: Given transformation matrix [a b; c d] and point (x, y): (1) Write point as column vector [x; y]. (2) Multiply: first row of matrix times vector gives x'-coordinate = a·x + b·y. Second row times vector gives y'-coordinate = c·x + d·y. (3) Result is transformed point (x', y') = (ax+by, cx+dy). Example: [0 -1; 1 0] applied to (5, 3): x' = 0·5 + (-1)·3 = -3, y' = 1·5 + 0·3 = 5, so image is (-3, 5). That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form [cos(θ) -sin(θ); sin(θ) cos(θ)]—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: [0 -1; 1 0] is 90° rotation, [-1 0; 0 -1] is 180° rotation. REFLECTION matrices have form [±1 0; 0 ±1] with exactly one negative—[1 0; 0 -1] reflects across x-axis (y negated), [-1 0; 0 1] reflects across y-axis (x negated). SCALING matrices have equal diagonal entries [k 0; 0 k]—both coordinates multiplied by same k, or different entries [a 0; 0 b] for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). Well done spotting the rotation matrix!

Question 4

Two transformation matrices A=(2101)A = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} and $$B = \begin{pmatrix} 1 & 0 \ -1 & 2 \end{pmatrix}

  1. 1414 units² because both matrices have the same determinant value (correct answer)
  2. 77 units² because matrix B has half the area scaling effect of matrix A
  3. 2828 units² because matrix B has twice the area scaling effect of matrix A
  4. 2121 units² because the area scaling effects have a 3:2 ratio favoring matrix B
Explanation: det(A) = 2×1 - 1×0 = 2, and det(B) = 1×2 - 0×(-1) = 2. Both have the same area scaling factor of 2, so they produce equal areas when applied to the same original region. Choices B, C, and D incorrectly calculate the determinants or their relationship.

Question 5

Triangle ABCABC has vertices A(1,1)A(1,1), B(3,1)B(3,1), and C(2,4)C(2,4). It is transformed by the matrix K=(2002).K=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}. What are the coordinates of C′C' (the image of CC)?​

  1. (4,4)(4,4)
  2. (4,8)(4,8) (correct answer)
  3. (2,8)(2,8)
  4. (1,2)(1,2)
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. The matrix K = [2 0; 0 2] represents uniform scaling by factor 2. Applying this to vertex C(2, 4): K[2; 4] = [2 0; 0 2][2; 4] = [2·2 + 0·4; 0·2 + 2·4] = [4 + 0; 0 + 8] = [4; 8], giving C'(4, 8). Choice B correctly shows this result where both coordinates are doubled: x: 2 → 4 and y: 4 → 8. Choice A incorrectly has y' = 4 (not doubled), choice C incorrectly has x' = 2 (not doubled), and choice D shows the original point divided by 2 instead of multiplied. Matrix multiplication for transformations: SCALING matrices have equal diagonal entries [k 0; 0 k]—both coordinates multiplied by same k. The scaling transformation K = [2 0; 0 2] doubles all distances from the origin, enlarging the triangle by factor 2!

Question 6

Triangle △PQR\triangle PQR has vertices P(1,1)P(1,1), Q(3,1)Q(3,1), and R(2,3)R(2,3). A transformation is given by T=(2002).T=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}. What are the coordinates of the image triangle △P′Q′R′\triangle P'Q'R'?

  1. P′(2,2), Q′(6,2), R′(4,6)P'(2,2),\ Q'(6,2),\ R'(4,6) (correct answer)
  2. P′(2,2), Q′(5,2), R′(4,5)P'(2,2),\ Q'(5,2),\ R'(4,5)
  3. P′(12,12), Q′(32,12), R′(1,32)P'(\tfrac{1}{2},\tfrac{1}{2}),\ Q'(\tfrac{3}{2},\tfrac{1}{2}),\ R'(1,\tfrac{3}{2})
  4. P′(−2,−2), Q′(−6,−2), R′(−4,−6)P'(-2,-2),\ Q'(-6,-2),\ R'(-4,-6)
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. The matrix T = [2 0; 0 2] uniformly scales by factor 2, so we apply it to each vertex: P'= T[1; 1] = [2·1; 2·1] = [2; 2], Q'= T[3; 1] = [2·3; 2·1] = [6; 2], R'= T[2; 3] = [2·2; 2·3] = [4; 6]. Choice A correctly shows P'(2,2), Q'(6,2), R'(4,6) as the scaled triangle with all distances from origin doubled. The transformation preserves the triangle's shape while doubling its size—each coordinate is multiplied by 2, making the image triangle similar to the original with scale factor 2. Matrix multiplication for transformations: Uniform scaling [2 0; 0 2] multiplies both coordinates by 2. For triangle vertices: P(1,1)→P'(2,2), Q(3,1)→Q'(6,2), R(2,3)→R'(4,6). The scaling preserves angles and ratios of distances, creating a similar triangle that's twice as large!

Question 7

A transformation matrix T=(2103)T = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} is applied to a triangle with vertices at (0,0)(0,0), (2,0)(2,0), and (0,1)(0,1). What is the ratio of the area of the transformed triangle to the original triangle?

  1. 3:13:1
  2. 5:15:1
  3. 6:16:1 (correct answer)
  4. 7:17:1
Explanation: The ratio of areas under a linear transformation equals the absolute value of the determinant of the transformation matrix. For matrix T, det(T) = (2)(3) - (1)(0) = 6. Therefore, the area ratio is 6:1. Choice A uses only one diagonal element, choice B adds the matrix elements, and choice D uses the sum of all elements.

Question 8

Triangle ABCABC has vertices A(1,1)A(1,1), B(3,1)B(3,1), and C(2,3)C(2,3) on the coordinate plane. The triangle is transformed by S=(2002)S = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}. Which set of vertices gives the image triangle A′B′C′A'B'C'?

  1. A′(2,2),B′(6,2),C′(4,6)A'(2,2), B'(6,2), C'(4,6) (correct answer)
  2. A′(2,2),B′(5,2),C′(4,5)A'(2,2), B'(5,2), C'(4,5)
  3. A′(12,12),B′(32,12),C′(1,32)A'(\tfrac{1}{2},\tfrac{1}{2}), B'(\tfrac{3}{2},\tfrac{1}{2}), C'(1,\tfrac{3}{2})
  4. A′(−2,−2),B′(−6,−2),C′(−4,−6)A'(-2,-2), B'(-6,-2), C'(-4,-6)
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x;y][x; y] and multiply by transformation matrix T=[a b;c d]T = [a\ b; c\ d] using matrix multiplication: T[x;y]=[a b;c d][x;y]=[ax+by;cx+dy]=[x′;y′]T[x; y] = [a\ b; c\ d][x; y] = [ax + by; cx + dy] = [x'; y'] where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = [cos⁡(θ)−sin⁡(θ);sin⁡(θ) cos⁡(θ)][\cos(\theta) -\sin(\theta); \sin(\theta)\ \cos(\theta)] (example: 90° rotation uses θ=90° giving [0 −1;1 0][0\ -1; 1\ 0] since cos⁡(90°)=0\cos(90°)=0 and sin⁡(90°)=1\sin(90°)=1), REFLECTION across x-axis = [1 0;0 −1][1\ 0; 0\ -1] (keeps x same, negates y), REFLECTION across y-axis = [−1 0;0 1][-1\ 0; 0\ 1] (negates x, keeps y same), SCALING by factor k = [k 0;0 k][k\ 0; 0\ k] (multiplies both coordinates by k, enlarges by factor k). Applying S=[2 0;0 2]S=[2\ 0; 0\ 2] to each vertex: A(1,1)→(2,2), B(3,1)→(6,2), C(2,3)→(4,6). Choice A correctly scales all vertices by factor 2. Distractors like choice C might use scaling by 1/2 instead, shrinking the triangle. Transform each point individually and plot to verify the shape enlarges proportionally—great job, you're getting stronger!

Question 9

A transformation sends (xy)↦(x′y′)=(abcd)(xy). \begin{pmatrix}x\\ y\end{pmatrix}\mapsto \begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}. For the matrix M=(0110),M=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}, which statement best describes the transformation?​

  1. It reflects points across the line y=xy=x (swaps xx and yy). (correct answer)
  2. It reflects points across the xx-axis (negates yy).
  3. It rotates points 90∘90^\circ counterclockwise about the origin.
  4. It scales all points by factor 22.
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. For matrix M = [0 1; 1 0], applying to point (x, y): M[x; y] = [0 1; 1 0][x; y] = [0·x + 1·y; 1·x + 0·y] = [y; x], which swaps the x and y coordinates—this is exactly reflection across the line y = x. Choice A correctly identifies this as reflection across the line y = x, where each point (x, y) maps to (y, x). Choice B (reflection across x-axis) would need [1 0; 0 -1], C (90° rotation) would need [0 -1; 1 0], and D (scaling by 2) would need [2 0; 0 2]. The matrix [0 1; 1 0] has 1's on the anti-diagonal and 0's on the main diagonal, creating the coordinate swap (x, y) → (y, x). This reflection across y = x mirrors points across the 45° line through the origin where x equals y!

Question 10

The matrix R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} represents a rotation. A triangle undergoes this transformation followed by scaling matrix $$ \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}

  1. 1616 units² because rotation doubles area and scaling doubles it again
  2. 88 units² because the combined transformation has determinant 4 (correct answer)
  3. 44 units² because each transformation doubles the linear dimensions independently
  4. 3232 units² because rotation matrices preserve area completely in all compositions
Explanation: Rotation matrices have determinant 1 (preserve area), and the scaling matrix has determinant 4. Combined determinant is 1×4=4, so original area = 32/4 = 8. Choice A incorrectly claims rotation doubles area, choice C confuses linear and area scaling, choice D ignores the scaling effect.

Question 11

A shear transformation S=(1k01)S = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} is applied to a parallelogram with base 4 and height 3. After transformation, one vertex moves from (2,1)(2,1) to (5,1)(5,1). What is the area of the transformed parallelogram?

  1. 1212 units² because shear transformations preserve all area measurements (correct answer)
  2. 1818 units² because the base increases proportionally with the vertex displacement
  3. 1515 units² because the area scales by the amount of horizontal shear applied
  4. 2121 units² because both base and height change under the shear transformation
Explanation: Shear transformations have determinant 1, so they preserve area regardless of the shear parameter k. The original area 4×3=12 remains unchanged. The vertex displacement helps find k=3, but doesn't affect area preservation. Choices B, C, and D incorrectly assume shear transformations change area.

Question 12

Which transformation is represented by the matrix $$T=\begin{pmatrix}1&0\0&-1\end{pmatrix}

  1. Reflection across the yy-axis
  2. Rotation 90∘90^\circ counterclockwise about the origin
  3. Reflection across the xx-axis (correct answer)
  4. Scaling by factor −1-1 in the xx-direction only
Explanation: This question tests your ability to use 2×22 \times 2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×22 \times 2 matrix can represent a linear transformation of the plane: to transform a point (x,y)(x, y), write it as column vector [x;y][x; y] and multiply by transformation matrix T=(abcd)T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} using matrix multiplication: T[x;y]=(abcd)[x;y]=[ax+by;cx+dy]=[x′;y′]T[x; y] = \begin{pmatrix} a & b \\ c & d \end{pmatrix} [x; y] = [ax+by; cx+dy] = [x'; y'] where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ\theta counterclockwise = (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} (example: 90∘90^\circ rotation uses θ=90∘\theta=90^\circ giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} since cos⁡(90∘)=0\cos(90^\circ)=0 and sin⁡(90∘)=1\sin(90^\circ)=1), REFLECTION across x-axis = (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (keeps x same, negates y), REFLECTION across y-axis = (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (negates x, keeps y same), SCALING by factor k = (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! Here, T=(100−1)T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} applied to a general (x,y)(x, y) gives (x,−y)(x, -y), which keeps x the same and negates y, matching a reflection across the x-axis. Choice C correctly identifies this as reflection across the x-axis based on the matrix entries where the y-component is negated while x remains unchanged. A distractor like choice A might confuse it with y-axis reflection, which would have (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} instead, so always check which diagonal entry is negative to distinguish axis reflections. Matrix multiplication for transformations: Given transformation matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} and point (x,y)(x, y): (1) Write point as column vector [x;y][x; y]. (2) Multiply: first row of matrix times vector gives x'-coordinate = a⋅x+b⋅ya \cdot x + b \cdot y. Second row times vector gives y'-coordinate = c⋅x+d⋅yc \cdot x + d \cdot y. (3) Result is transformed point (x′,y′)=(ax+by,cx+dy)(x', y') = (ax+by, cx+dy). Example: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} applied to (5,3)(5, 3): x' = 0⋅5+(−1)⋅3=−30 \cdot 5 + (-1) \cdot 3 = -3, y' = 1⋅5+0⋅3=51 \cdot 5 + 0 \cdot 3 = 5, so image is (−3,5)(-3, 5). That's a 90∘90^\circ counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} is 90∘90^\circ rotation, (−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} is 180∘180^\circ rotation. REFLECTION matrices have form (±100±1)\begin{pmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{pmatrix} with exactly one negative—(100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} reflects across x-axis (y negated), (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} reflects across y-axis (x negated). SCALING matrices have equal diagonal entries (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}—both coordinates multiplied by same k, or different entries (a00b)\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). You're doing great—keep matching matrix patterns to transformation types!

Question 13

A transformation matrix is M=(2001).M=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}. What does MM do to a point (x,y)(x, y)?

  1. Doubles yy and keeps xx the same
  2. Reflects across the xx-axis
  3. Doubles xx and keeps yy the same (correct answer)
  4. Scales both coordinates by 22
Explanation: This question tests your ability to use 2×22 \times 2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×22 \times 2 matrix can represent a linear transformation of the plane: to transform a point (x,y)(x, y), write it as column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} and multiply by transformation matrix T=(abcd)T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} using matrix multiplication: T(xy)=(abcd)(xy)=(ax+bycx+dy)=(x′y′)T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ\theta counterclockwise =(cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))= \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} (example: 90∘90^\circ rotation uses θ=90∘\theta=90^\circ giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} since cos⁡(90∘)=0\cos(90^\circ)=0 and sin⁡(90∘)=1\sin(90^\circ)=1), REFLECTION across x-axis =(100−1)= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (keeps x same, negates y), REFLECTION across y-axis =(−1001)= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (negates x, keeps y same), SCALING by factor k =(k00k)= \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)⋅\cdot(rotation matrix). The resulting product matrix represents the combined transformation in one step! The matrix M=(2001)M = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} transforms (x,y)(x, y) to (2x,y)(2x, y), doubling the x-coordinate while leaving y unchanged, which is a non-uniform scaling in the x-direction. Choice C correctly captures this effect by analyzing the diagonal entries, where 2 scales x and 1 keeps y the same. Distractors like choice A might reverse the coordinates affected, but remember the top-left entry scales x and bottom-right scales y in diagonal matrices. Matrix multiplication for transformations: Given transformation matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} and point (x,y)(x, y): (1) Write point as column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}. (2) Multiply: first row of matrix times vector gives x'-coordinate =a⋅x+b⋅y= a \cdot x + b \cdot y. Second row times vector gives y'-coordinate =c⋅x+d⋅y= c \cdot x + d \cdot y. (3) Result is transformed point (x′,y′)=(ax+by,cx+dy)(x', y') = (ax + by, cx + dy). Example: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} applied to (5,3)(5, 3): x′=0⋅5+(−1)⋅3=−3x' = 0 \cdot 5 + (-1) \cdot 3 = -3, y′=1⋅5+0⋅3=5y' = 1 \cdot 5 + 0 \cdot 3 = 5, so image is (−3,5)(-3, 5). That's a 90∘90^\circ counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} is 90∘90^\circ rotation, (−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} is 180∘180^\circ rotation. REFLECTION matrices have form (±100±1)\begin{pmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{pmatrix} with exactly one negative—(100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} reflects across x-axis (y negated), (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} reflects across y-axis (x negated). SCALING matrices have equal diagonal entries (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}—both coordinates multiplied by same k, or different entries (a00b)\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). You're building strong skills in non-uniform transformations—keep going!

Question 14

A scaling transformation multiplies both coordinates by 33. Which matrix represents this transformation?

  1. [1003]\begin{bmatrix}1&0\\0&3\end{bmatrix}
  2. [3003]\begin{bmatrix}3&0\\0&3\end{bmatrix} (correct answer)
  3. [0330]\begin{bmatrix}0&3\\3&0\end{bmatrix}
  4. [3001]\begin{bmatrix}3&0\\0&1\end{bmatrix}
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = [cos(θ) -sin(θ); sin(θ) cos(θ)] (example: 90° rotation uses θ=90° giving [0 -1; 1 0] since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = [1 0; 0 -1] (keeps x same, negates y), REFLECTION across y-axis = [-1 0; 0 1] (negates x, keeps y same), SCALING by factor k = [k 0; 0 k] (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! Scaling both coordinates by 3 means multiplying x and y by 3, represented by the diagonal matrix with 3 on both diagonals. Choice B correctly identifies [3 0; 0 3] as the uniform scaling matrix by factor 3. A distractor like choice A might result from confusing uniform scaling with non-uniform, where only one coordinate is scaled by 3. Matrix multiplication for transformations: Given transformation matrix [a b; c d] and point (x, y): (1) Write point as column vector [x; y]. (2) Multiply: first row of matrix times vector gives x'-coordinate = a·x + b·y. Second row times vector gives y'-coordinate = c·x + d·y. (3) Result is transformed point (x', y') = (ax+by, cx+dy). Example: [0 -1; 1 0] applied to (5, 3): x' = 0·5 + (-1)·3 = -3, y' = 1·5 + 0·3 = 5, so image is (-3, 5). That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form [cos(θ) -sin(θ); sin(θ) cos(θ)]—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: [0 -1; 1 0] is 90° rotation, [-1 0; 0 -1] is 180° rotation. REFLECTION matrices have form [±1 0; 0 ±1] with exactly one negative—[1 0; 0 -1] reflects across x-axis (y negated), [-1 0; 0 1] reflects across y-axis (x negated). SCALING matrices have equal diagonal entries [k 0; 0 k]—both coordinates multiplied by same k, or different entries [a 0; 0 b] for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). Excellent work recognizing scaling matrices!

Question 15

Let R=(0−110)R=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix} (rotate 90∘90^\circ counterclockwise) and F=(100−1)F=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} (reflect across the xx-axis). What is the matrix for "rotate first, then reflect," i.e., F RF\,R?

  1. (0−110)\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. (0110)\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. To find "rotate first, then reflect," we compute F·R = [1 0; 0 -1]·[0 -1; 1 0] = [1·0 + 0·1, 1·(-1) + 0·0; 0·0 + (-1)·1, 0·(-1) + (-1)·0] = [0, -1; -1, 0], giving the matrix [0 -1; -1 0]. Choice B correctly shows [0 -1; -1 0] as the composition of rotating 90° counterclockwise then reflecting across the x-axis. This combined transformation takes (1, 0) to (0, 1) after rotation, then to (0, -1) after reflection; and takes (0, 1) to (-1, 0) after rotation, which stays at (-1, 0) after x-axis reflection. Matrix multiplication for transformations: For composition F·R, multiply [1 0; 0 -1]·[0 -1; 1 0]. First column: [1 0; 0 -1]·[0; 1] = [0; -1]. Second column: [1 0; 0 -1]·[-1; 0] = [-1; 0]. Result: [0 -1; -1 0]. The order matters—R·F would give a different result!

Question 16

Let R=(0−110)R=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix} (rotate 90∘90^\circ counterclockwise) and S=(2002)S=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix} (scale by 22). A point is first rotated by RR and then scaled by SS. What is the combined transformation matrix?​

  1. (0−220)\begin{pmatrix}0 & -2\\ 2 & 0\end{pmatrix} (correct answer)
  2. (0−110)\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}
  3. (0−2−20)\begin{pmatrix}0 & -2\\ -2 & 0\end{pmatrix}
  4. (0−120)\begin{pmatrix}0 & -1\\ 2 & 0\end{pmatrix}
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. For composition of transformations, we multiply matrices in reverse order: if we first apply R then S, the combined matrix is S·R. Computing S·R = [2 0; 0 2]·[0 -1; 1 0] = [2·0 + 0·1, 2·(-1) + 0·0; 0·0 + 2·1, 0·(-1) + 2·0] = [0, -2; 2, 0] = [0 -2; 2 0]. Choice A correctly shows this combined transformation matrix [0 -2; 2 0]. Choice B shows just the rotation matrix without scaling, choice C has wrong sign in bottom-left entry, and choice D has inconsistent scaling factors. COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The combined matrix [0 -2; 2 0] rotates 90° counterclockwise AND scales by 2 in one step!

Question 17

A point S(3,2)S(3,2) is transformed by the matrix T=(1201).T=\begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}. What are the coordinates of S′S'?

  1. (3,8)(3,8)
  2. (7,2)(7,2) (correct answer)
  3. (5,2)(5,2)
  4. (7,5)(7,5)
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = [cos(θ) -sin(θ); sin(θ) cos(θ)] (example: 90° rotation uses θ=90° giving [0 -1; 1 0] since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = [1 0; 0 -1] (keeps x same, negates y), REFLECTION across y-axis = [-1 0; 0 1] (negates x, keeps y same), SCALING by factor k = [k 0; 0 k] (multiplies both coordinates by k, enlarges by factor k). For S(3,2) and T=[1 2; 0 1], S' = [13 + 22; 03 + 12] = [7; 2], so (7,2)—this is a shear transformation. Choice B correctly applies the matrix multiplication to get (7,2). Distractors like choice A might ignore the off-diagonal term, giving (3,8) from a misread. Even for non-standard matrices, follow the same multiplication steps and check calculations—keep going, you're improving!

Question 18

A transformation is defined by A=[2001].A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}. What does this transformation do to a point (x,y)(x, y)?

  1. Reflects across the xx-axis
  2. Doubles the xx-coordinate and leaves the yy-coordinate unchanged (correct answer)
  3. Doubles the yy-coordinate and leaves the xx-coordinate unchanged
  4. Scales both coordinates by 2
Explanation: This question tests your ability to use 2×22 \times 2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×22 \times 2 matrix can represent a linear transformation of the plane: to transform a point (x,y)(x, y), write it as column vector [xy]\begin{bmatrix} x \\ y \end{bmatrix} and multiply by transformation matrix T=[abcd]T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} using matrix multiplication: T[xy]=[abcd][xy]=[ax+bycx+dy]=[x′y′]T \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a x + b y \\ c x + d y \end{bmatrix} = \begin{bmatrix} x' \\ y' \end{bmatrix} where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ\theta counterclockwise = [cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ)]\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} (example: 90° rotation uses θ=90°\theta=90° giving [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} since cos⁡(90°)=0\cos(90°)=0 and sin⁡(90°)=1\sin(90°)=1), REFLECTION across x-axis = [100−1]\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} (keeps x same, negates y), REFLECTION across y-axis = [−1001]\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} (negates x, keeps y same), SCALING by factor k = [k00k]\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! The matrix A=[2001]A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} transforms (x,y)(x, y) to (2x,y)(2x, y), doubling the x-coordinate while leaving y unchanged. Choice B correctly describes this non-uniform scaling effect by analyzing the diagonal entries. A distractor like choice D might confuse it with uniform scaling, which would require 2 on both diagonals. Matrix multiplication for transformations: Given transformation matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} and point (x,y)(x, y): (1) Write point as column vector [xy]\begin{bmatrix} x \\ y \end{bmatrix}. (2) Multiply: first row of matrix times vector gives x'-coordinate = a⋅x+b⋅ya \cdot x + b \cdot y. Second row times vector gives y'-coordinate = c⋅x+d⋅yc \cdot x + d \cdot y. (3) Result is transformed point (x′,y′)=(ax+by,cx+dy)(x', y') = (a x + b y, c x + d y). Example: [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} applied to (5,3)(5, 3): x' = 0⋅5+(−1)⋅3=−30 \cdot 5 + (-1) \cdot 3 = -3, y' = 1⋅5+0⋅3=51 \cdot 5 + 0 \cdot 3 = 5, so image is (−3,5)(-3, 5). That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form [cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ)]\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}—look for this pattern with cos and -sin in first row, sin and cos in second row. Common: [0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} is 90° rotation, [−100−1]\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} is 180° rotation. REFLECTION matrices have form [±100±1]\begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{bmatrix} with exactly one negative—[100−1]\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} reflects across x-axis (y negated), [−1001]\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} reflects across y-axis (x negated). SCALING matrices have equal diagonal entries [k00k]\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}—both coordinates multiplied by same k, or different entries [a00b]\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). You're building strong skills in matrix identification!

Question 19

A point P(4,3)P(4,3) is transformed by the matrix T=(0−110)T=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} . What are the coordinates of the image point P′P' after applying TT (using (x′y′)=T(xy)\begin{pmatrix} x' \\ y' \end{pmatrix} = T \begin{pmatrix} x \\ y \end{pmatrix})?

  1. (−4,−3)(-4,-3)
  2. (3,−4)(3,-4)
  3. (−3,4)(-3,4) (correct answer)
  4. (4,−3)(4,-3)
Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} and multiply by transformation matrix T = (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} using matrix multiplication: T(xy)=(abcd)(xy)=(ax+bycx+dy)=(x′y′)T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(θ) & -\sin(θ) \\ \sin(θ) & \cos(θ) \end{pmatrix} (example: 90° rotation uses θ=90° giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (keeps x same, negates y), REFLECTION across y-axis = (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (negates x, keeps y same), SCALING by factor k = (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} (multiplies both coordinates by k, enlarges by factor k). For point P(4,3) and T=(0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, compute P' = (0×4+(−1)×31×4+0×3)=(−34)\begin{pmatrix} 0 \times 4 + (-1) \times 3 \\ 1 \times 4 + 0 \times 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}, so (-3,4). Choice C correctly applies the matrix multiplication to get (-3,4), matching the 90° counterclockwise rotation that swaps and signs coordinates appropriately. A common distractor like choice A might result from swapping signs incorrectly or misapplying the matrix rows, leading to (-4,-3) instead. Keep practicing matrix multiplication for transformations: always multiply the matrix by the column vector, and verify by checking if the result aligns with the geometric effect, like preserving distance for rotations—you've got this!

Question 20

First rotate a point 90∘90^\circ counterclockwise about the origin, then scale by a factor of 22. Let R=(0−110),S=(2002).R=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix},\quad S=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}. What single matrix represents the combined transformation (apply RR first, then SS)?

  1. (0−220)\begin{pmatrix}0 & -2\\ 2 & 0\end{pmatrix} (correct answer)
  2. (0−110)\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}
  3. (0−120)\begin{pmatrix}0 & -1\\ 2 & 0\end{pmatrix}
  4. (0−210)\begin{pmatrix}0 & -2\\ 1 & 0\end{pmatrix}
Explanation: This question tests your ability to use 2×22 \times 2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×22 \times 2 matrix can represent a linear transformation of the plane: to transform a point (x,y)(x, y), write it as column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} and multiply by transformation matrix T=(abcd)T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} using matrix multiplication: T(xy)=(abcd)(xy)=(ax+bycx+dy)=(x′y′)T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax+by \\ cx+dy \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} where (x′,y′)(x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ\theta counterclockwise = (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} (example: 90° rotation uses θ=90∘\theta=90^\circ giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} since cos⁡(90∘)=0\cos(90^\circ)=0 and sin⁡(90∘)=1\sin(90^\circ)=1), REFLECTION across x-axis = (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (keeps x same, negates y), REFLECTION across y-axis = (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (negates x, keeps y same), SCALING by factor k = (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix) = S∗R=(2002)∗(0−110)=(0−220)S * R = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} * \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}. Choice A correctly computes this combined matrix for rotation first then scaling. Distractors like choice B might ignore the composition and just give the rotation matrix. Remember to multiply in the order of application (second times first)—practice with test points to verify, and you'll be a pro!