Solutions — Geometric Transformations with Matrices

Q1 — Applying a Rotation Matrix

R(90°) = [[0,−1],[1,0]]

[[0,−1],[1,0]] [[4],[−2]] = [[0(4)+(−1)(−2)],[1(4)+0(−2)]] = [[2],[4]]

Image: (2, 4)

Check: |original| = √(16+4) = √20, |image| = √(4+16) = √20 ✓

Q2 — Reflection in the x-axis

Reflection in x-axis: [[1,0],[0,−1]] maps (x,y) to (x,−y).

(a) (3,5) → (3,−5)

(b) (−2,4) → (−2,−4)

(c) (0,−7) → (0,7)

Q3 — Dilation Matrix

Dilation matrix: [[3,0],[0,3]].

A′: [[3,0],[0,3]][[1],[0]] = [[3],[0]] = (3,0)

B′: [[3,0],[0,3]][[2],[1]] = [[6],[3]] = (6,3)

C′: [[3,0],[0,3]][[0],[2]] = [[0],[6]] = (0,6)

Area of original triangle = ½|base × height| = ½(2)(2) = 2. Image area = 9 × 2 = 18 square units (scale factor for area = k² = 9).

Q4 — Rotation by 45°

R(45°) = [[cos45°, −sin45°],[sin45°, cos45°]] = [[(1/√2), −(1/√2)],[(1/√2), (1/√2)]]

Image of (2, 0): [[(1/√2)(2)],[(1/√2)(2)]] = [[ √2],[√2]]

Image: (√2, √2)

Check: rotating (2,0) by 45° gives distance 2 at angle 45° = (2cos45°, 2sin45°) = (√2, √2) ✓

Q5 — Identifying Transformations from Matrices

(a) [[0,1],[1,0]]: maps (x,y) to (y,x). This is a reflection in y = x. (det = 0−1 = −1, confirms reflection.)

(b) [[−1,0],[0,−1]]: maps (x,y) to (−x,−y). This is a rotation by 180°. (det = −1×−1 = 1, confirms rotation/orientation-preserving.)

(c) [[2,0],[0,1]]: maps (x,y) to (2x,y). This is a horizontal dilation by factor 2 (x-direction only). (det = 2, areas doubled.)

Q6 — Composition of Transformations

T₁ = R(90°) = [[0,−1],[1,0]], T₂ = reflection in x-axis = [[1,0],[0,−1]]

Composite = T₂T₁ (T₁ applied first, goes on the right):

[[1,0],[0,−1]] [[0,−1],[1,0]]

Entry (1,1): 1(0)+0(1) = 0 Entry (1,2): 1(−1)+0(0) = −1

Entry (2,1): 0(0)+(−1)(1) = −1 Entry (2,2): 0(−1)+(−1)(0) = 0

Composite = [[0,−1],[−1,0]] (reflection in y = −x)

Apply to (1,2): [[0,−1],[−1,0]][[1],[2]] = [[(0)(1)+(−1)(2)],[(−1)(1)+(0)(2)]] = [[−2],[−1]]

Image: (−2, −1)

Q7 — Area Scale Factor

The area scale factor for a linear transformation is |det(A)|.

A = [[2, 1], [0, 3]], so det(A) = 2(3) − 1(0) = 6.

Original area = 1 (unit square).

Image area = |det(A)| × original area = 6 × 1 = 6 square units

Verification by finding image vertices:

(0,0) → [[2,1],[0,3]][[0],[0]] = (0,0)

(1,0) → [[2,1],[0,3]][[1],[0]] = (2,0)

(1,1) → [[2,1],[0,3]][[1],[1]] = (3,3)

(0,1) → [[2,1],[0,3]][[0],[1]] = (1,3)

Parallelogram with sides (2,0) and (1,3): area = |2×3 − 0×1| = 6 ✓

Q8 — Finding the Matrix for a Reflection in y = −x

Reflecting (1,0) in y = −x: the reflection sends (a,b) to (−b,−a). So (1,0) → (0,−1).

Reflecting (0,1) in y = −x: (0,1) → (−1,0).

Matrix: columns are images of basis vectors. Matrix = [[0,−1],[−1,0]]

Apply to (3,−2): [[0,−1],[−1,0]][[3],[−2]] = [[(0)(3)+(−1)(−2)],[(−1)(3)+(0)(−2)]] = [[2],[−3]]

Image: (2,−3)

Check: midpoint of (3,−2) and (2,−3) is (2.5,−2.5), which lies on y=−x ✓

Gradient joining them: (−3−(−2))/(2−3) = (−1)/(−1) = 1, perpendicular to y=−x (gradient −1) ✓

Q9 — Two Reflections = One Rotation

T₁ = reflection in x-axis = [[1,0],[0,−1]], T₂ = reflection in y-axis = [[−1,0],[0,1]].

Composite T₂T₁ (x-axis first, so T₁ on right):

[[−1,0],[0,1]][[1,0],[0,−1]]

(1,1): (−1)(1)+0(0) = −1 (1,2): (−1)(0)+0(−1) = 0

(2,1): 0(1)+1(0) = 0 (2,2): 0(0)+1(−1) = −1

Composite = [[−1,0],[0,−1]] = R(180°) ✓

Apply to (2,5): [[−1,0],[0,−1]][[2],[5]] = [[ −2],[−5]] = (−2, −5)

Key result: composing two reflections gives a rotation. det(T₂T₁) = (−1)(−1) = 1 (rotation). The angle of rotation = twice the angle between the mirror lines (x-axis and y-axis are 90° apart → rotation by 180°).

Q10 — Finding the Angle of Rotation from a Matrix

The rotation matrix maps (1,0) to (cosθ, sinθ). We are told this image is (1/2, √3/2).

So cosθ = 1/2 and sinθ = √3/2, giving θ = 60° = π/3.

The rotation matrix is R(60°) = [[1/2, −√3/2],[ √3/2, 1/2]].

Apply to (0, 4):

x′ = (1/2)(0) + (−√3/2)(4) = −2√3

y′ = (√3/2)(0) + (1/2)(4) = 2

Image: (−2√3, 2)

Check distance: |(−2√3, 2)| = √(12+4) = √16 = 4 = |(0,4)| ✓