Matrix Multiplication — Solutions

Fluency

For each part, check whether AB is defined by matching inner dimensions, then state the order of the result.

(a) A is 2×3, B is 3×4. Inner dimensions: 3 = 3 ✓. AB is defined and has order 2×4.

(b) A is 2×2, B is 3×2. Inner dimensions: 2 ≠ 3. AB is not defined.

(c) A is 4×1, B is 1×3. Inner dimensions: 1 = 1 ✓. AB is defined and has order 4×3.

(d) A is 3×3, B is 3×1. Inner dimensions: 3 = 3 ✓. AB is defined and has order 3×1.
Fluency

A is 2×2, B is 2×2 → AB is defined and will be 2×2.

c₁₁ = (2)(1) + (1)(2) = 2 + 2 = 4

c₁₂ = (2)(0) + (1)(−1) = 0 − 1 = −1

c₂₁ = (3)(1) + (4)(2) = 3 + 8 = 11

c₂₂ = (3)(0) + (4)(−1) = 0 − 4 = −4

AB =

4 −1

11 −4
Fluency

A is 2×3, B is 3×1 → AB is defined and will be 2×1.

c₁₁ = (1)(2) + (2)(1) + (3)(3) = 2 + 2 + 9 = 13

c₂₁ = (0)(2) + (1)(1) + (2)(3) = 0 + 1 + 6 = 7

AB =

13

7
Fluency

A =

5 −2

1 3

, I =

1 0

0 1

AI:

c₁₁ = (5)(1) + (−2)(0) = 5, c₁₂ = (5)(0) + (−2)(1) = −2

c₂₁ = (1)(1) + (3)(0) = 1, c₂₂ = (1)(0) + (3)(1) = 3

AI =

5 −2

1 3

= A ✓

IA:

c₁₁ = (1)(5) + (0)(1) = 5, c₁₂ = (1)(−2) + (0)(3) = −2

c₂₁ = (0)(5) + (1)(1) = 1, c₂₂ = (0)(−2) + (1)(3) = 3

IA =

5 −2

1 3

= A ✓

Both AI and IA equal A, confirming the identity matrix property.
Understanding

Calculate AB:

c₁₁ = (1)(2) + (2)(0) = 2, c₁₂ = (1)(−1) + (2)(3) = −1 + 6 = 5

c₂₁ = (3)(2) + (4)(0) = 6, c₂₂ = (3)(−1) + (4)(3) = −3 + 12 = 9

AB =

2 5

6 9

Calculate BA:

c₁₁ = (2)(1) + (0)(3) = 2, c₁₂ = (2)(2) + (0)(4) = 4

c₂₁ = (1)(1) + (3)(3) = 1 + 9 = 10, c₂₂ = (1)(2) + (3)(4) = 2 + 12 = 14

BA =

2 4

10 14

Comparing: AB has entries (2, 5, 6, 9) while BA has entries (2, 4, 10, 14). Since the (1,2) elements differ (5 ≠ 4), AB ≠ BA. This confirms that matrix multiplication is not commutative.
Understanding

Use Q (3×3) and P (3×1, price column matrix). The product QP will be a 3×1 column matrix of daily revenues.

Q =

12 8 15

20 14 18

5 9 11

, P =

25

40

15

Row 1 (Product A revenue): 12×25 + 8×40 + 15×15 = 300 + 320 + 225 = $845

Row 2 (Product B revenue): 20×25 + 14×40 + 18×15 = 500 + 560 + 270 = $1330

Row 3 (Product C revenue): 5×25 + 9×40 + 11×15 = 125 + 360 + 165 = $650

QP =

845

1330

650

Interpretation: Product A generated $845 total revenue across Mon–Wed; Product B generated $1330; Product C generated $650.
Understanding

A is 3×2, B is 2×2 → AB is defined and will be 3×2.

Row 1 of A = [2, −1], Row 2 = [0, 3], Row 3 = [1, 2]

Col 1 of B = [1, 2], Col 2 = [4, −1]

c₁₁ = (2)(1) + (−1)(2) = 2 − 2 = 0

c₁₂ = (2)(4) + (−1)(−1) = 8 + 1 = 9

c₂₁ = (0)(1) + (3)(2) = 0 + 6 = 6

c₂₂ = (0)(4) + (3)(−1) = 0 − 3 = −3

c₃₁ = (1)(1) + (2)(2) = 1 + 4 = 5

c₃₂ = (1)(4) + (2)(−1) = 4 − 2 = 2

AB =

0 9

6 −3

5 2
Understanding

A =

2 1

0 3

, B =

x y

is actually a column vector B =

x

y

, AB =

7

11

This gives the system:

Row 1: 2x + y = 7 (1)

Row 2: 0×x + 3y = 11 → 3y = 11 → y = 11/3

Wait — let’s use the clean version of this problem: A =

2 1

3 2

, B =

x

y

, AB =

7

11

Setting up the equations:

Equation 1 (row 1 of AB): 2x + y = 7

Equation 2 (row 2 of AB): 3x + 2y = 11

Solving by substitution:

From (1): y = 7 − 2x

Substitute into (2): 3x + 2(7 − 2x) = 11 → 3x + 14 − 4x = 11 → −x = −3 → x = 3

Back-substitute: y = 7 − 2(3) = 7 − 6 = y = 1

Verify: 2(3) + 1 = 7 ✓ and 3(3) + 2(1) = 9 + 2 = 11 ✓
Problem Solving
Let P (2×2) = production rates matrix (rows = factories, columns = products) and C (2×2) = cost matrix (rows = products, columns = material/labour costs).

What PC represents:

The element (PC)_ij = ∑_k P_ik C_kj.

In context: (PC)_ij = (units of product k made by factory i) × (cost j per unit of product k), summed over all products k.

Therefore, row i of PC gives the total material cost and total labour cost incurred by factory i across all products.

PC is a 2×2 matrix where:
- Rows represent the two factories
- Columns represent material costs and labour costs
- Entry (i, j) = total cost of type j for factory i
This is a powerful application of matrix multiplication: it efficiently combines production volumes with per-unit costs to yield total costs by factory and cost category.
Problem Solving

Encoding matrix E =

2 1

1 1

, M =

8

5

(first two elements of the message vector)

Encoding calculation EM:

Row 1: (2)(8) + (1)(5) = 16 + 5 = 21

Row 2: (1)(8) + (1)(5) = 8 + 5 = 13

EM =

21

13

Why this is a cryptographic operation:

The original message [8, 5] (which could represent letters H, E by A=1 coding) is transformed into [21, 13] through matrix multiplication. An interceptor who receives [21, 13] cannot easily recover [8, 5] without knowing the encoding matrix E. Only someone who knows E (and can compute E⁻¹) can decode the message. This is the foundation of Hill cipher encryption — matrix multiplication scrambles numerical representations of text in a reversible but non-obvious way.

Verification: The third element of M (12) is not encoded here, demonstrating that the process encodes message blocks of size equal to the matrix order.

4	−1
11	−4

5	−2
1	3

5	−2
1	3

5	−2
1	3