Matrices and applications (WACE Year 12 Mathematics Applications)

Syllabus dot point

“Perform matrix operations, find determinants and inverses of 2x2 matrices, solve matrix equations, and apply transition matrices to model systems.”

9 min readUpdated 2 Jun 2026

Quick answer

A matrix is a rectangular array of numbers. You add and subtract matrices of the same size entry by entry, multiply by matching the rows of the first to the columns of the second, and for a $2\times2$ matrix find the determinant $ad - bc$ and the inverse by swapping the diagonal, negating the off-diagonal and dividing by the determinant. The inverse solves matrix equations, and a transition matrix multiplied by a state vector models how a system changes step by step.

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What this dot point is asking
Matrix operations
Determinant and inverse of a 2x2 matrix
Solving matrix equations
Transition matrices
Reaching the steady state
Exam-style questions

What this dot point is asking

You must carry out matrix operations, find determinants and inverses of $2\times2$ matrices, solve matrix equations, and apply transition matrices.

Matrix operations

Matrices are added, subtracted and scaled entry by entry, but multiplication has its own rule.

Matrix multiplication

To find the product $AB$ , the number of columns of $A$ must equal the number of rows of $B$ . Each entry of the product is the sum of products of a row of $A$ with a column of $B$ . Matrix multiplication is not commutative: usually $AB \neq BA$ .

For example, $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 1(5) + 2(6) \\ 3(5) + 4(6) \end{pmatrix} = \begin{pmatrix} 17 \\ 39 \end{pmatrix}$ .

Determinant and inverse of a 2x2 matrix

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

Determinant and inverse

$\det A = ad - bc$ .

$A^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ , provided $\det A \neq 0$ .

The recipe for the inverse is: swap the entries on the leading diagonal, negate the other two, and divide every entry by the determinant. If $\det A = 0$ the matrix is singular and has no inverse.

Solving matrix equations

A system of two linear equations can be written as $A\mathbf{x} = \mathbf{b}$ , where $A$ holds the coefficients, $\mathbf{x}$ the unknowns and $\mathbf{b}$ the constants. Pre-multiply both sides by $A^{-1}$ to get $\mathbf{x} = A^{-1}\mathbf{b}$ .

Pre-multiply, do not divide

There is no matrix division. To solve $A\mathbf{x} = \mathbf{b}$ , multiply on the left by $A^{-1}$ , giving $\mathbf{x} = A^{-1}\mathbf{b}$ . Because multiplication is not commutative, the inverse must go on the left of $\mathbf{b}$ .

Transition matrices

A transition matrix $T$ describes how a system moves between states each step. Each column gives the proportions moving from one state to each state, so its columns sum to $1$ . The state after one step is $S_{n+1} = T S_n$ , and after $n$ steps $S_n = T^n S_0$ .

Worked example

A transition matrix over several steps

In a town, each year $90\%$ of city residents stay in the city and $10\%$ move to the suburbs, while $20\%$ of suburb residents move to the city and $80\%$ stay. Start with $6000$ in the city and $4000$ in the suburbs.

Write the transition matrix

With the city as the top row and column, $T = \begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix}$ , and $S_0 = \begin{pmatrix} 6000 \\ 4000 \end{pmatrix}$ .

After one year

$S_1 = T S_0 = \begin{pmatrix} 0.9(6000) + 0.2(4000) \\ 0.1(6000) + 0.8(4000) \end{pmatrix} = \begin{pmatrix} 5400 + 800 \\ 600 + 3200 \end{pmatrix} = \begin{pmatrix} 6200 \\ 3800 \end{pmatrix}$ .

After two years

$S_2 = T S_1 = \begin{pmatrix} 0.9(6200) + 0.2(3800) \\ 0.1(6200) + 0.8(3800) \end{pmatrix} = \begin{pmatrix} 5580 + 760 \\ 620 + 3040 \end{pmatrix} = \begin{pmatrix} 6340 \\ 3660 \end{pmatrix}$ .

The steady state

Repeated multiplication settles towards a steady state where $S_{n+1} = S_n$ . Here the long-run distribution approaches about $6667$ city and $3333$ suburb residents, the proportions that no longer change from year to year.

Reaching the steady state

For a regular transition matrix the state vector converges to a steady state independent of where it started, found either by iterating $T^n S_0$ for large $n$ on the calculator, or by solving $T \mathbf{s} = \mathbf{s}$ with the entries of $\mathbf{s}$ summing to the total population.

Common errors

Multiplying in the wrong order: $AB \neq BA$ ; for transition models always compute $T S_n$ , with $T$ on the left.
Forgetting to divide by the determinant: The inverse is the swapped-negated matrix divided by $\det A$ ; skipping the division gives a matrix that is not the inverse.
Inverting a singular matrix: If $\det A = 0$ , no inverse exists and the system has no unique solution.
Columns of a transition matrix not summing to 1: Each column must total $1$ (every member of a state goes somewhere); a column that does not is a setup error.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks

Let

A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}

. (a) Find the determinant of

A

. (b) Find

A^{-1}

. (c) Hence solve

A\mathbf{x} = \begin{pmatrix} 8 \\ 6 \end{pmatrix}

Show worked answer →

Use the $2\times2$ determinant and inverse formulas, then multiply.

(a) $\det A = (3)(4) - (2)(1) = 12 - 2 = 10$ . (1 mark)

(b) $A^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & -2 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.2 \\ -0.1 & 0.3 \end{pmatrix}$ . (2 marks)

(c) $\mathbf{x} = A^{-1}\begin{pmatrix} 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 0.4(8) - 0.2(6) \\ -0.1(8) + 0.3(6) \end{pmatrix} = \begin{pmatrix} 3.2 - 1.2 \\ -0.8 + 1.8 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ . (3 marks)

Markers reward the determinant, the swap-negate-divide inverse, and pre-multiplying by $A^{-1}$ to solve.

WACE 2023-style6 marks

A transition matrix

T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}

describes the movement of customers between Brand A (top row) and Brand B each month. The current state is

S_0 = \begin{pmatrix} 500 \\ 500 \end{pmatrix}

. (a) Find the distribution after one month. (b) Find it after two months.

Show worked answer →

Multiply the transition matrix by the state vector each month.

(a) $S_1 = T S_0 = \begin{pmatrix} 0.8(500) + 0.3(500) \\ 0.2(500) + 0.7(500) \end{pmatrix} = \begin{pmatrix} 400 + 150 \\ 100 + 350 \end{pmatrix} = \begin{pmatrix} 550 \\ 450 \end{pmatrix}$ . (3 marks)

(b) $S_2 = T S_1 = \begin{pmatrix} 0.8(550) + 0.3(450) \\ 0.2(550) + 0.7(450) \end{pmatrix} = \begin{pmatrix} 440 + 135 \\ 110 + 315 \end{pmatrix} = \begin{pmatrix} 575 \\ 425 \end{pmatrix}$ . (3 marks)

Markers reward $S_{n+1} = T S_n$ applied in order, with columns of $T$ summing to $1$ .