What is 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.

SCSA WACE 2021-style practice: Let A = pmatrix 3 & 2 \\ 1 & 4 pmatrix. (a) Find the determinant of A. (b) Find A^-1. (c) Hence solve Ax = pmatrix 8 \\ 6 pmatrix.

Use the 22 determinant and inverse formulas, then multiply. (a) A = (3)(4) - (2)(1) = 12 - 2 = 10. (1 mark) (b) A^-1 = 110pmatrix 4 & -2 \\ -1 & 3 pmatrix = pmatrix 0.4 & -0.2 \\ -0.1 & 0.3 pmatrix. (2 marks) (c) x = A^-1pmatrix 8 \\ 6 pmatrix = pmatrix 0.4(8) - 0.2(6) \\ -0.1(8) + 0.3(6) pmatrix = pmatrix 3.2 - 1.2 \\ -0.8 + 1.8 pmatrix = pmatrix 2 \\ 1 pmatrix. (3 marks) Markers reward the determinant, the swap-negate-divide inverse, and pre-multiplying by A^-1 to solve.

§-Syllabus dot point

WAMathematics ApplicationsSyllabus dot point

How can a grid of numbers store data and model change over time?

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

How to add, multiply and invert matrices, solve matrix equations with the inverse, and use transition matrices and steady states to model populations and market share.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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

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 $\mathbf{x} = A^{-1}\mathbf{b}$ .

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksLet

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 20236 marksA 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 $1$ .