Matrices — Topic Review — Solutions

15 questions covering all Matrices sub-topics: notation, operations, multiplication, determinants, inverses and solving systems. Click Show Solution to reveal full working.

Fluency

Matrix M =

1	5	−3
4	0	2

. (a) State the order of M. (b) Identify element m₁₂. (c) Identify element m₂₃.

(a) M has 2 rows and 3 columns: order 2 × 3

(b) m₁₂ = element in row 1, column 2 = 5

(c) m₂₃ = element in row 2, column 3 = 2

Fluency
Calculate 2A − B where A =

3 −1

0 2

and B =

1 4

−2 1

2A =

6 −2

0 4

2A − B =

6−1 −2−4

0−(−2) 4−1

=

5 −6

2 3
Fluency
Find AB where A =

2 1

0 3

and B =

1 2

4 −1

Row 1 of A times columns of B:
c₁₁ = 2(1)+1(4) = 6 c₁₂ = 2(2)+1(−1) = 3
c₂₁ = 0(1)+3(4) = 12 c₂₂ = 0(2)+3(−1) = −3

AB =

6 3

12 −3
Fluency
Find det(A) for A =

5 2

3 1

det(A) = 5×1 − 2×3 = 5 − 6 = −1
Fluency
Find A⁻¹ for A =

3 1

5 2

det(A) = 3×2 − 1×5 = 6 − 5 = 1

A⁻¹ = (1/1)

2 −1

−5 3
Understanding
Explain why the matrix

2 4

1 2

has no inverse. What does this mean when using the matrix method to solve a system?

det = 2×2 − 4×1 = 4 − 4 = 0. A matrix with determinant 0 is singular and has no inverse. When this is the coefficient matrix for a system of equations, the matrix method cannot be used — the system either has no solution (inconsistent parallel lines) or infinitely many solutions (coincident lines).
Understanding
Find the value of k such that

k 3

2 k

is singular.

For a singular matrix, det = 0: k×k − 3×2 = 0 → k² − 6 = 0 → k² = 6 → k = ±√6 ≈ ±2.449
Understanding
Solve using the matrix inverse method:
3x + y = 7
x + y = 3

A =

3 1

1 1

, B =

7

3

det(A) = 3 − 1 = 2. A⁻¹ = ½

1 −1

−1 3

X = ½

7−3

−7+9

= ½

4

2

=

2

1

x = 2, y = 1
Understanding
A 2×2 matrix P has det(P) = 4. What is det(2P)?

For a 2×2 matrix, det(kP) = k² × det(P).

det(2P) = 2² × 4 = 4 × 4 = 16

Reason: Each element of P is multiplied by 2. The determinant formula ad − bc contains each element exactly once, but since there are 2 elements being scaled in each product, the overall determinant is multiplied by k² = 4.
Understanding
A store tracks stock of 2 products across 2 branches. Opening stock matrix S =

30 20

15 25

. Sales matrix =

12 8

5 10

. Find closing stock. Rows = Products A, B. Columns = Branches 1, 2.

Closing stock = S − Sales =

30−12 20−8

15−5 25−10

=

18 12

10 15

Product A: Branch 1 has 18, Branch 2 has 12. Product B: Branch 1 has 10, Branch 2 has 15.
Understanding
Find AB and BA for A =

1 0

2 1

and B =

2 1

0 3

. Is AB = BA?

AB: c₁₁=2, c₁₂=1, c₂₁=4+0=4, c₂₂=2+3=5 →

2 1

4 5

BA: c₁₁=2+2=4, c₁₂=0+1=1, c₂₁=0+6=6, c₂₂=0+3=3 →

4 1

6 3

AB ≠ BA — matrix multiplication is not commutative in general.
Problem Solving
Solve using the matrix method, then verify by substitution:
2x − y = 3
x + 2y = 9

A =

2 −1

1 2

, B =

3

9

det(A) = 4 − (−1) = 5

A⁻¹ = ⅕

2 1

−1 2

X = ⅕

6+9

−3+18

= ⅕

15

15

=

3

3

x = 3, y = 3. Verify: 2(3)−3=3 ✓ and 3+2(3)=9 ✓
Problem Solving
An event sells adult tickets ($12) and child tickets ($7). 80 tickets were sold for $760. Set up a matrix equation and solve to find how many of each ticket type were sold.

Let a = adult tickets, c = child tickets.

System: a + c = 80 and 12a + 7c = 760

A =

1 1

12 7

det(A) = 7 − 12 = −5

A⁻¹ = −⅕

7 −1

−12 1

X = −⅕

560 − 760

−960 + 760

= −⅕

−200

−200

=

40

40

40 adult and 40 child tickets. Check: 40+40=80 ✓ and 12(40)+7(40)=480+280=760 ✓
Problem Solving
Given A =

a 2

1 3

and det(A) = 4, find a. Then write down A⁻¹.

det(A) = 3a − 2(1) = 3a − 2 = 4 → 3a = 6 → a = 2

A =

2 2

1 3

A⁻¹ = ¼

3 −2

−1 2
Problem Solving
A coding matrix E =

1 1

0 1

encodes message M =

5

3

. (a) Find the encoded message EM. (b) The inverse of E is E⁻¹ =

1 −1

0 1

. Decode the message

9

4

.

(a) EM =

1 1

0 1

5

3

=

5+3

0+3

=

8

3

(b) Apply E⁻¹ to the encoded message:

1 −1

0 1

9

4

=

9−4

0+4

=

5

4

Decoded message is [5, 4].

3	−1
0	2

1	4
−2	1

6	−2
0	4

6−1	−2−4
0−(−2)	4−1