Solving Systems of Equations Using Matrices

Key Terms

Matrix equation: A system of linear equations can be written as AX = B, where A = coefficient matrix, X = unknowns column, B = constants column.
Solution by inverse: X = A⁻¹B (valid only when det(A) ≠ 0).
Unique solution: det(A) ≠ 0 → exactly one solution.
No solution: det(A) = 0 and the system is inconsistent → the lines are parallel (no intersection).
Infinite solutions: det(A) = 0 and the system is consistent → the equations represent the same line.
Row reduction: An alternative method: form the augmented matrix [A|B] and apply elementary row operations to find the solution.

The Matrix Method: AX = B

Setting up: Write the system as a matrix equation AX = B
where A = coefficient matrix, X = variable column vector, B = constant column vector.

Solving: X = A⁻¹B (multiply both sides on the left by A⁻¹)

Condition: Only works when det(A) ≠ 0 (A must be invertible).

For 2×2: A = [[a,b],[c,d]] → A⁻¹ = (1/det A) [[d, −b], [−c, a]]
det(A) = ad − bc (QCAA Formula Sheet)

Steps for the Matrix Method

Step	Action
1	Write each equation in the form ax + by = c
2	Identify coefficient matrix A, variable vector X, constant vector B
3	Calculate det(A) = ad − bc. If det = 0, stop — no unique solution.
4	Find A⁻¹ using the formula
5	Calculate X = A⁻¹B by matrix multiplication
6	Write the solution and verify by substituting back

Worked Example 1 — Solving a 2×2 System

Solve: 2x + y = 7 and x + 3y = 11

Step 1–2: Set up AX = B

A =

2	1
1	3

, X =

, B =

Step 3: det(A) = 2×3 − 1×1 = 6 − 1 = 5 ≠ 0 → unique solution exists.

Step 4: A⁻¹ = ⅕

3	−1
−1	2

Step 5: X = A⁻¹B = ⅕

3×7 − 1×11

−1×7 + 2×11

= ⅕

Solution: x = 2, y = 3. Check: 2(2)+3=7 ✓ and 2+3(3)=11 ✓

Worked Example 2 — Practical Application

A farmer buys x bags of fertiliser at $18 each and y bags of seed at $24 each. He buys 10 bags total and spends $204. Find x and y.

System: x + y = 10 and 18x + 24y = 204

A =

1	1
18	24

, B =

204

det(A) = 24 − 18 = 6. A⁻¹ = ⅙

24	−1
−18	1

X = A⁻¹B = ⅙

240 − 204

−180 + 204

= ⅙

Solution: 6 bags of fertiliser, 4 bags of seed.

Hot Tip: Always check det(A) before calculating the inverse. If det = 0 the system is either inconsistent (no solution) or dependent (infinitely many solutions) — you cannot use the matrix method. The presence of a non-zero determinant guarantees a unique solution.

Full Lesson: Solving Systems Using Matrices

Why Matrix Methods?

In Unit 1 you solved simultaneous equations using substitution and elimination. The matrix method is a powerful alternative that extends naturally to larger systems (3 equations, 3 unknowns; 4 equations, 4 unknowns; and so on). For 2×2 systems in General Maths, the process always follows the same structured steps.

The Matrix Equation AX = B

Any system of two linear equations can be written as a single matrix equation:

System: ax + by = e and cx + dy = f

Matrix form: A · X = B where A =

a	b
c	d

, X =

, B =

Why X = A⁻¹B (not B·A⁻¹)?

From AX = B, we multiply both sides on the left by A⁻¹:

A⁻¹(AX) = A⁻¹B → (A⁻¹A)X = A⁻¹B → IX = A⁻¹B → X = A⁻¹B

Matrix multiplication is NOT commutative, so the order matters. A⁻¹B and BA⁻¹ give different results.

Connection to the Real World

Businesses use matrix equations constantly: resource allocation (how many of each product to manufacture given cost and time constraints), logistics (how many items to ship via each route), finance (portfolio balancing). In each case, setting up AX = B and solving X = A⁻¹B gives the answer directly.

What If det(A) = 0?

A determinant of zero means the two equations are either parallel lines (no intersection → no solution) or the same line (coincident → infinitely many solutions). In either case, no unique solution exists and the matrix method cannot be applied.

Lesson Tip: When setting up the matrix equation, make sure both equations are in the standard form ax + by = c before extracting the coefficients. If an equation has variables on the right side (e.g., 3x = 5 − 2y), rearrange it first.

Mastery Practice

See Answers ➔

Fluency
Write the system in matrix form AX = B (identify A, X, and B only — do not solve):
3x + 2y = 11
x − y = 1
Fluency
Solve the system using the matrix inverse method:
2x + y = 7
x + y = 5
Fluency
Solve the system using the matrix inverse method:
4x − y = 9
2x + y = 3
Fluency
Solve using the matrix method:
x + 3y = 10
2x − y = 1
Understanding
The coefficient matrix for a system is A =

1 2

2 4

. Without solving, explain what happens when you attempt to use the matrix inverse method, and what this tells you about the system of equations geometrically.
Understanding
Solve the system and verify your answer by substitution:
3x + 5y = 1
x + 2y = 0
Understanding
A café sells lattes for $5 and cappuccinos for $4. On Monday 40 drinks were sold for $182. Set up and solve a matrix equation to find the number of each type sold.
Understanding
Solve the following system using the matrix inverse method. Show all steps including finding det(A) and A⁻¹.
2x − 3y = 4
x + y = 3
Problem Solving
A sports club sells adult tickets ($15) and concession tickets ($9). A total of 120 tickets are sold for $1470.
1. Write a system of two equations, then express it in matrix form AX = B.
2. Find A⁻¹ and hence solve for the number of each ticket type sold.
3. Verify your solution is correct.
Problem Solving
Two alloys are made from copper (Cu) and zinc (Zn). Alloy X is 60% Cu and 40% Zn. Alloy Y is 30% Cu and 70% Zn. A metalworker wants to combine x kg of Alloy X with y kg of Alloy Y to produce 100 kg of a mix that is 45% copper. Set up and solve the system using the matrix method.