Matrices and Systems of Equations — Topic Review — Solutions

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This review covers all four lessons in the Matrices and Systems of Equations topic: Matrix Operations and Properties, Determinants and Inverses, Geometric Transformations with Matrices, and Solving Systems with Matrices. Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.

Mixed Review Questions

1. Fluency

   Q1 — Matrix Addition and Scalar Multiplication

   Let A = [[2, −1], [3, 4]] and B = [[0, 5], [−2, 1]]. Calculate: (a) A + B    (b) 3A    (c) 2A − B

   ▶ Show Answer

   (a) A + B = [[2+0, −1+5], [3+(−2), 4+1]] = [[2, 4], [1, 5]]

   (b) 3A = [[3×2, 3×(−1)], [3×3, 3×4]] = [[6, −3], [9, 12]]

   (c) 2A = [[4, −2], [6, 8]]

   2A − B = [[4−0, −2−5], [6−(−2), 8−1]] = [[4, −7], [8, 7]]

2. Fluency

   Q2 — Matrix Multiplication

   Let P = [[1, 2], [0, 3]] and Q = [[4, −1], [2, 5]]. Calculate PQ and QP. Are they equal?

   ▶ Show Answer

   PQ:

   Row 1: [1×4 + 2×2,   1×(−1) + 2×5] = [8, 9]

   Row 2: [0×4 + 3×2,   0×(−1) + 3×5] = [6, 15]

   PQ = [[8, 9], [6, 15]]

   QP:

   Row 1: [4×1 + (−1)×0,   4×2 + (−1)×3] = [4, 5]

   Row 2: [2×1 + 5×0,   2×2 + 5×3] = [2, 19]

   QP = [[4, 5], [2, 19]]

   PQ ≠ QP. Matrix multiplication is not commutative in general.

3. Fluency

   Q3 — Determinant of a 2×2 Matrix

   Find the determinant of each matrix: (a) A = [[5, 2], [3, 1]]    (b) B = [[−4, 6], [2, −3]]

   ▶ Show Answer

   For a 2×2 matrix [[a, b], [c, d]], det = ad − bc.

   (a) det(A) = 5×1 − 2×3 = 5 − 6 = −1

   (b) det(B) = (−4)×(−3) − 6×2 = 12 − 12 = 0

   Since det(B) = 0, matrix B is singular (has no inverse).

4. Fluency

   Q4 — Inverse of a 2×2 Matrix

   Find the inverse of A = [[3, 1], [5, 2]].

   ▶ Show Answer

   For A = [[a, b], [c, d]], A⁻¹ = 1/det(A) × [[d, −b], [−c, a]].

   det(A) = 3×2 − 1×5 = 6 − 5 = 1

   A⁻¹ = 1/1 × [[2, −1], [−5, 3]] = [[2, −1], [−5, 3]]

   Verify: AA⁻¹ = [[3×2+1×(−5), 3×(−1)+1×3], [5×2+2×(−5), 5×(−1)+2×3]] = [[1, 0], [0, 1]] ✓

5. Fluency

   Q5 — Reflection Matrix

   State the transformation matrix for a reflection in the x-axis, and apply it to the point (3, −2).

   ▶ Show Answer

   The reflection matrix for the x-axis is:

   M = [[1, 0], [0, −1]]

   Apply to point (3, −2) — write as column vector [3, −2]^T:

   M × [[3], [−2]] = [[1×3 + 0×(−2)], [0×3 + (−1)×(−2)]] = [[3], [2]]

   The image of (3, −2) under reflection in the x-axis is (3, 2).

6. Understanding

   Q6 — Properties of Matrix Operations

   Given A = [[1, 0], [2, −1]], verify that (A²)^T = (A^T)². Is this always true?

   ▶ Show Answer

   Calculate A²:

   A² = A × A = [[1,0],[2,−1]] × [[1,0],[2,−1]]

   = [[1+0, 0+0], [2−2, 0+1]] = [[1, 0], [0, 1]]

   (A²)^T = [[1, 0], [0, 1]]^T = [[1, 0], [0, 1]]

   Calculate A^T:

   A^T = [[1, 2], [0, −1]]

   (A^T)² = [[1,2],[0,−1]] × [[1,2],[0,−1]] = [[1+0, 2−2],[0+0, 0+1]] = [[1, 0], [0, 1]]

   (A²)^T = [[1,0],[0,1]] = (A^T)² ✓

   Note: In general, (AB)^T = B^TA^T, so (A²)^T = (AA)^T = A^TA^T = (A^T)². This is always true.

7. Understanding

   Q7 — Determinant of a 3×3 Matrix

   Calculate det(C) where C = [[2, 1, 0], [3, −1, 2], [1, 4, −1]].

   ▶ Show Answer

   Expanding along the first row:

   det(C) = 2 × det([[−1, 2],[4, −1]]) − 1 × det([[3, 2],[1, −1]]) + 0 × det([[3, −1],[1, 4]])

   = 2[((−1)(−1) − 2×4)] − 1[(3(−1) − 2×1)] + 0

   = 2[1 − 8] − 1[−3 − 2]

   = 2(−7) − 1(−5)

   = −14 + 5

   det(C) = −9

8. Understanding

   Q8 — Rotation Matrix

   A rotation of 90° anticlockwise about the origin is applied to the triangle with vertices A(1, 0), B(3, 0), C(3, 2). Find the images A′, B′, C′.

   ▶ Show Answer

   The rotation matrix for angle θ anticlockwise is R = [[cosθ, −sinθ], [sinθ, cosθ]].

   For θ = 90°: R = [[0, −1], [1, 0]].

   A′: [[0,−1],[1,0]] × [[1],[0]] = [[0],[1]]  →  A′(0, 1)

   B′: [[0,−1],[1,0]] × [[3],[0]] = [[0],[3]]  →  B′(0, 3)

   C′: [[0,−1],[1,0]] × [[3],[2]] = [[−2],[3]]  →  C′(−2, 3)

9. Understanding

   Q9 — Solving a System with Inverse Matrices

   Solve the system using the inverse matrix method:
   2x + 3y = 7
   x + 2y = 4

   ▶ Show Answer

   Write as AX = B where A = [[2, 3], [1, 2]], X = [[x], [y]], B = [[7], [4]].

   Find A⁻¹:

   det(A) = 2×2 − 3×1 = 4 − 3 = 1

   A⁻¹ = 1/1 × [[2, −3], [−1, 2]] = [[2, −3], [−1, 2]]

   Solve X = A⁻¹B:

   [[x],[y]] = [[2,−3],[−1,2]] × [[7],[4]]

   = [[2×7 + (−3)×4], [(−1)×7 + 2×4]]

   = [[14 − 12], [−7 + 8]] = [[2], [1]]

   x = 2, y = 1

   Verify: 2(2) + 3(1) = 7 ✓   and   2 + 2(1) = 4 ✓

10. Understanding

    Q10 — Dilation Matrix

    Describe the geometric transformation given by T = [[3, 0], [0, 3]], and find the image of the line y = 2x − 1 under this transformation.

    ▶ Show Answer

    T = [[3,0],[0,3]] = 3I is a dilation from the origin with scale factor 3. Every point is scaled to three times its distance from the origin in the same direction.

    If (x, y) maps to (x′, y′), then x′ = 3x and y′ = 3y, so x = x′/3 and y = y′/3.

    Substitute into y = 2x − 1:

    y′/3 = 2(x′/3) − 1

    y′ = 2x′ − 3

    The image of the line is y = 2x − 3 (same gradient, y-intercept scaled by factor 3).

11. Problem Solving

    Q11 — Solving a 3×3 System by Row Reduction

    Solve the system using row reduction (Gaussian elimination):
    x + 2y − z = 3
    2x − y + z = 1
    3x + y + 2z = 7

    ▶ Show Answer

    Write the augmented matrix [A|B] and row reduce:

    [[1, 2, −1, | 3], [2, −1, 1, | 1], [3, 1, 2, | 7]]

    R2 → R2 − 2R1:   [0, −5, 3, | −5]

    R3 → R3 − 3R1:   [0, −5, 5, | −2]

    [[1, 2, −1, | 3], [0, −5, 3, | −5], [0, −5, 5, | −2]]

    R3 → R3 − R2:   [0, 0, 2, | 3]

    [[1, 2, −1, | 3], [0, −5, 3, | −5], [0, 0, 2, | 3]]

    Back-substitute:

    From R3: 2z = 3  →  z = 3/2

    From R2: −5y + 3(3/2) = −5  →  −5y = −5 − 9/2 = −19/2  →  y = 19/10

    From R1: x + 2(19/10) − 3/2 = 3  →  x = 3 − 19/5 + 3/2 = 30/10 − 38/10 + 15/10 = 7/10  →  x = 7/10

    Solution: x = 7/10, y = 19/10, z = 3/2

12. Problem Solving

    Q12 — Composite Transformations

    Find the single matrix that represents: first a reflection in the y-axis, then a rotation of 90° anticlockwise about the origin. Apply this composite to the point (2, 1).

    ▶ Show Answer

    Reflection in the y-axis: M_y = [[−1, 0], [0, 1]]

    Rotation 90° anticlockwise: R₉₀ = [[0, −1], [1, 0]]

    Composite (rotation after reflection): C = R₉₀ × M_y

    C = [[0, −1], [1, 0]] × [[−1, 0], [0, 1]]

    = [[0×(−1) + (−1)×0,   0×0 + (−1)×1], [1×(−1) + 0×0,   1×0 + 0×1]]

    = [[0, −1], [−1, 0]]

    This is a reflection in the line y = −x.

    Apply to (2, 1):

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

    Image: (−1, −2)

13. Problem Solving

    Q13 — Singular Systems and Geometric Interpretation

    Analyse the system 2x + 4y = 6 and x + 2y = k for values of k that give (a) no solution (b) infinitely many solutions. Explain geometrically.

    ▶ Show Answer

    Write the augmented matrix: [[2, 4, | 6], [1, 2, | k]]

    R1 → R1/2: [[1, 2, | 3], [1, 2, | k]]

    R2 → R2 − R1: [[1, 2, | 3], [0, 0, | k−3]]

    (a) No solution when k − 3 ≠ 0, i.e., k ≠ 3.

    The second equation becomes 0 = k−3 (a contradiction).

    Geometrically: both lines have the same gradient (−1/2) but different y-intercepts — they are parallel and never intersect.

    (b) Infinitely many solutions when k = 3.

    The system reduces to x + 2y = 3 — both equations represent the same line.

    General solution: let y = t (free parameter), then x = 3 − 2t.

    Geometrically: the two lines are identical — they coincide.

14. Problem Solving

    Q14 — Finding a Matrix from Transformation Properties

    A linear transformation T maps (1, 0) to (3, 1) and (0, 1) to (−2, 4). Find the matrix of T and the image of the point (5, −3).

    ▶ Show Answer

    For a linear transformation represented by matrix M = [[a, b], [c, d]]:

    M × [[1],[0]] = [[a],[c]] = [[3],[1]]  →  a = 3, c = 1

    M × [[0],[1]] = [[b],[d]] = [[−2],[4]]  →  b = −2, d = 4

    M = [[3, −2], [1, 4]]

    Image of (5, −3):

    M × [[5],[−3]] = [[3×5 + (−2)×(−3)], [1×5 + 4×(−3)]] = [[15+6], [5−12]] = [[21], [−7]]

    Image: (21, −7)

15. Problem Solving

    Q15 — Network and Matrix Application

    Three siblings share pocket money. Each week: Alex gives 20% of his money to Beth and 10% to Cal; Beth gives 30% of her money to Alex and 10% to Cal; Cal gives 20% of her money to Alex and 30% to Beth. Set up the transition matrix and find each sibling’s share after one week if they start with $100, $80, $60 respectively.

    ▶ Show Answer

    The transition matrix T (columns represent outgoing fractions from each person, rows represent incoming to each person):

    Alex keeps 100% − 20% − 10% = 70% = 0.7

    Beth keeps 100% − 30% − 10% = 60% = 0.6

    Cal keeps 100% − 20% − 30% = 50% = 0.5

    T = [[0.7, 0.3, 0.2], [0.2, 0.6, 0.3], [0.1, 0.1, 0.5]]

    Column 1 (from Alex): Alex keeps 0.7, gives 0.2 to Beth, gives 0.1 to Cal.

    Column 2 (from Beth): Alex gets 0.3, Beth keeps 0.6, Cal gets 0.1.

    Column 3 (from Cal): Alex gets 0.2, Beth gets 0.3, Cal keeps 0.5.

    Multiply T × [100, 80, 60]^T:

    Alex: 0.7(100) + 0.3(80) + 0.2(60) = 70 + 24 + 12 = $106

    Beth: 0.2(100) + 0.6(80) + 0.3(60) = 20 + 48 + 18 = $86

    Cal: 0.1(100) + 0.1(80) + 0.5(60) = 10 + 8 + 30 = $48

    Total: 106 + 86 + 48 = $240 ✓ (same as 100 + 80 + 60)