Solutions — Matrix Operations and Properties

Q1 — Matrix Addition and Scalar Multiplication

A = [[3,−1],[2,4]], B = [[−2,5],[1,−3]]

(a) A + B: add each corresponding entry.

Row 1: 3+(−2)=1, −1+5=4. Row 2: 2+1=3, 4+(−3)=1.

A + B = [[1, 4], [3, 1]]

(b) 3A: multiply every entry of A by 3.

3A = [[9, −3], [6, 12]]

(c) 2A − B: first compute 2A = [[6,−2],[4,8]], then subtract B = [[−2,5],[1,−3]].

Row 1: 6−(−2)=8, −2−5=−7. Row 2: 4−1=3, 8−(−3)=11.

2A − B = [[8, −7], [3, 11]]

Q2 — Matrix Multiplication

A = [[1,2],[3,4]], B = [[5,6],[7,8]]. Compute each of the four entries of AB:

(1,1): row 1 of A · col 1 of B = 1(5) + 2(7) = 5 + 14 = 19

(1,2): row 1 of A · col 2 of B = 1(6) + 2(8) = 6 + 16 = 22

(2,1): row 2 of A · col 1 of B = 3(5) + 4(7) = 15 + 28 = 43

(2,2): row 2 of A · col 2 of B = 3(6) + 4(8) = 18 + 32 = 50

AB = [[19, 22], [43, 50]]

Q3 — Multiplying a Matrix by the Identity

A = [[5,−2],[3,7]], I = [[1,0],[0,1]].

Compute AI:

(1,1): 5(1)+(−2)(0)=5. (1,2): 5(0)+(−2)(1)=−2.

(2,1): 3(1)+7(0)=3. (2,2): 3(0)+7(1)=7.

AI = [[5,−2],[3,7]] = A ✓

Compute IA:

(1,1): 1(5)+0(3)=5. (1,2): 1(−2)+0(7)=−2.

(2,1): 0(5)+1(3)=3. (2,2): 0(−2)+1(7)=7.

IA = [[5,−2],[3,7]] = A ✓

Both products equal A, confirming AI = IA = A.

Q4 — Multiplying Non-Square Matrices

A is 2×3 and B is 3×1. The inner dimensions both equal 3, so AB is defined and has dimensions 2×1.

Row 1 of A · B: 1(7) + 2(8) + 3(9) = 7 + 16 + 27 = 50

Row 2 of A · B: 4(7) + 5(8) + 6(9) = 28 + 40 + 54 = 122

AB = [[50], [122]]

Note: BA would require B (3×1) × A (2×3), which has inner dimensions 1 ≠ 2, so BA is undefined.

Q5 — Showing AB ≠ BA

A = [[1,2],[0,1]], B = [[1,0],[3,1]].

Compute AB:

(1,1): 1(1)+2(3)=7. (1,2): 1(0)+2(1)=2. (2,1): 0(1)+1(3)=3. (2,2): 0(0)+1(1)=1.

AB = [[7,2],[3,1]]

Compute BA:

(1,1): 1(1)+0(0)=1. (1,2): 1(2)+0(1)=2. (2,1): 3(1)+1(0)=3. (2,2): 3(2)+1(1)=7.

BA = [[1,2],[3,7]]

Since [[7,2],[3,1]] ≠ [[1,2],[3,7]], we confirm AB ≠ BA. Matrix multiplication is not commutative.

Q6 — Finding Unknown Entries

[[x,2],[1,y]] [[3],[4]] = [[7],[5]]

Performing the multiplication on the left:

Row 1: x(3) + 2(4) = 3x + 8

Row 2: 1(3) + y(4) = 3 + 4y

Setting equal to the right-hand side:

Equation 1: 3x + 8 = 11 → 3x = 3 → x = 1

Equation 2: 3 + 4y = 7 → 4y = 4 → y = 1

Verify: [[1,2],[1,1]] [[3],[4]] = [[3+8],[3+4]] = [[11],[7]] ✓

Q7 — Transpose Property

A = [[1,3],[2,4]], B = [[0,1],[2,5]].

Compute AB:

(1,1): 0+6=6. (1,2): 1+15=16. (2,1): 0+8=8. (2,2): 2+20=22.

AB = [[6,16],[8,22]]. Therefore (AB)^T = [[6,8],[16,22]].

Compute B^TA^T:

A^T = [[1,2],[3,4]], B^T = [[0,2],[1,5]]

B^TA^T: (1,1): 0(1)+2(3)=6. (1,2): 0(2)+2(4)=8. (2,1): 1(1)+5(3)=16. (2,2): 1(2)+5(4)=22.

B^TA^T = [[6,8],[16,22]]

(AB)^T = B^TA^T = [[6,8],[16,22]] ✓

This confirms the transpose rule — note carefully that the order of A and B reverses in the formula.

Q8 — Setting Up a System as AX = B

System: 3x + 2y = 7 and −x + 4y = 5.

The coefficient matrix is formed by reading off the coefficients of x and y from each equation (left to right).

A = [[3, 2], [−1, 4]] (coefficient matrix)

X = [[x], [y]] (variable vector)

B = [[7], [5]] (right-hand side vector)

Verifying AX = B by expansion:

Row 1: 3x + 2y = 7 ✓

Row 2: −x + 4y = 5 ✓

This compact matrix notation is the foundation for solving linear systems using matrix inverses or Gaussian elimination.

Q9 — Powers of a Matrix

A = [[1,1],[0,1]].

A² = A·A:

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

A² = [[1,2],[0,1]]

A³ = A²·A:

(1,1): 1+0=1. (1,2): 1(1)+2(1)=3. (2,1): 0. (2,2): 1.

A³ = [[1,3],[0,1]]

Pattern and conjecture: The top-right entry equals the power n, while the diagonal remains 1 and the bottom-left remains 0.

Conjecture: Aⁿ = [[1,n],[0,1]] for all positive integers n.

This can be proved by mathematical induction (a perfect topic for Specialist Unit 1 Induction skills!).

Q10 — Proof: (A + B)² ≠ A² + 2AB + B² in General

A = [[1,1],[0,0]], B = [[0,0],[1,1]].

Left side: (A+B)²

A + B = [[1,1],[1,1]]. (A+B)²: (1,1)=1+1=2, (1,2)=1+1=2, (2,1)=1+1=2, (2,2)=1+1=2.

(A+B)² = [[2,2],[2,2]]

Right side: A² + 2AB + B²

A² = A·A: (1,1)=1, (1,2)=1, (2,1)=0, (2,2)=0. A² = [[1,1],[0,0]]

B² = B·B: (1,1)=1, (1,2)=1, (2,1)=1, (2,2)=1. B² = [[1,1],[1,1]]

AB: (1,1)=1, (1,2)=1, (2,1)=0, (2,2)=0. AB = [[1,1],[0,0]]

BA: (1,1)=0, (1,2)=0, (2,1)=1, (2,2)=1. BA = [[0,0],[1,1]]

A² + 2AB + B² = [[1,1],[0,0]] + 2[[1,1],[0,0]] + [[1,1],[1,1]] = [[4,4],[1,1]]

[[2,2],[2,2]] ≠ [[4,4],[1,1]], confirming the expansion fails.

Explanation: The correct expansion is (A+B)² = A² + AB + BA + B². Since AB ≠ BA in general, AB + BA ≠ 2AB, so the real-number formula A² + 2AB + B² does not apply to matrices. Non-commutativity is the root cause.