AQA AQA 2020-style practice: The matrix A = pmatrix 3 & 1 \\ 2 & 4 pmatrix. Find A^-1 and hence solve the simultaneous equations 3x + y = 5 and 2x + 4y = 10.

For A = pmatrix a & b \\ c & d pmatrix, A = ad - bc and A^-1 = (1)/( A)pmatrix d & -b \\ -c & a pmatrix. Here A = (3)(4) - (1)(2) = 12 - 2 = 10, so A^-1 = (1)/(10)pmatrix 4 & -1 \\ -2 & 3 pmatrix. The system is Apmatrix x \\ y pmatrix = pmatrix 5 \\ 10 pmatrix, so pmatrix x \\ y pmatrix = A^-1pmatrix 5 \\ 10 pmatrix = (1)/(10)pmatrix 4(5) - 1(10) \\ -2(5) + 3(10) pmatrix = (1)/(10)pmatrix 10 \\ 20 pmatrix = pmatrix 1 \\ 2 pmatrix. So x = 1, y = 2. Markers reward the determinant, the inverse, and the matrix-product solution.

EnglandFurther MathsSyllabus dot point

How do matrices represent transformations and systems of equations, and how do you invert them?

Matrix arithmetic, determinants, inverses of 2x2 and 3x3 matrices, matrices as linear transformations, invariant points and lines, and solving systems of linear equations.

A focused answer to the AQA A-Level Further Mathematics matrices content, covering matrix arithmetic, determinants, inverses of 2x2 and 3x3 matrices, matrices as linear transformations, invariant points and lines, and solving systems of linear equations.

Generated by Claude Opus 4.811 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 arithmetic
Determinants and inverses
Matrices as transformations
Invariant points and lines
Solving systems of equations

What this dot point is asking

AQA wants you to add, subtract and multiply matrices, find determinants and inverses of $2 \times 2$ and $3 \times 3$ matrices, interpret matrices as linear transformations of the plane, find invariant points and lines, and use the inverse matrix to solve systems of simultaneous linear equations.

Matrix arithmetic

Matrix multiplication is row by column and is not commutative, so in general $AB \neq BA$ . The identity matrix $I$ leaves any matrix unchanged.

Determinants and inverses

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ the determinant is $\det A = ad - bc$ and the inverse is $A^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

Matrices as transformations

A $2 \times 2$ matrix maps the plane linearly, fixing the origin. The columns are the images of the basis vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , which is the fastest way to identify a transformation.

Invariant points and lines

An invariant point satisfies $A\mathbf{x} = \mathbf{x}$ , so it is fixed by the transformation; solving this gives a system whose only solution is usually the origin unless the matrix has special structure. An invariant line maps onto itself as a whole, though individual points on it may slide along it. To find invariant lines through the origin, look for directions $\mathbf{x}$ with $A\mathbf{x}$ parallel to $\mathbf{x}$ .

Solving systems of equations

A system of linear equations can be written as $A\mathbf{x} = \mathbf{b}$ . If $A$ is invertible, the unique solution is $\mathbf{x} = A^{-1}\mathbf{b}$ . If $\det A = 0$ the system has either no solutions or infinitely many.

Solve a system with the inverse

Solve $\begin{cases} 2x + y = 5 \\ x + 3y = 10 \end{cases}$ using a matrix inverse.

Step 1: Write the system in matrix form

Any pair of simultaneous linear equations can be expressed as $A\mathbf{x} = \mathbf{b}$ , where $A$ holds the coefficients, $\mathbf{x}$ is the column of unknowns, and $\mathbf{b}$ is the column of constants. This structure allows us to solve by inverting $A$ .

A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}, \qquad \mathbf{b} = \begin{pmatrix} 5 \\ 10 \end{pmatrix}

Step 2: Find the determinant and the inverse

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is $ad - bc$ . The inverse swaps the diagonal entries, changes the sign of the off-diagonal entries, and divides by the determinant. A non-zero determinant confirms a unique solution exists.

\det A = (2)(3) - (1)(1) = 6 - 1 = 5

A^{-1} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}

Step 3: Multiply $A^{-1}$ by $\mathbf{b}$ to find $\mathbf{x}$

The unique solution is $\mathbf{x} = A^{-1}\mathbf{b}$ , because multiplying both sides of $A\mathbf{x} = \mathbf{b}$ on the left by $A^{-1}$ gives $\mathbf{x}$ directly. Carry out the matrix-vector multiplication.

\mathbf{x} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 15 - 10 \\ -5 + 20 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 5 \\ 15 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}

Final answer: $x = 1$ , $y = 3$ .

For a $3\times 3$ matrix the same ideas extend: the determinant is found by cofactor expansion along a row, a non-zero determinant guarantees an inverse, and the inverse solves a three-variable system. A zero determinant signals a singular matrix where the equations are either inconsistent (no solution) or dependent (infinitely many), which links directly to the geometry of three planes that fail to meet at a single point.

Exam-style practice questions

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

AQA 20206 marksThe matrix

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

. Find

A^{-1}

and hence solve the simultaneous equations

3x + y = 5

and

2x + 4y = 10

Show worked answer →

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , $\det A = ad - bc$ and $A^{-1} = \frac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

Here $\det A = (3)(4) - (1)(2) = 12 - 2 = 10$ , so $A^{-1} = \frac{1}{10}\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix}$ .

The system is $A\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \end{pmatrix}$ , so $\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 4(5) - 1(10) \\ -2(5) + 3(10) \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 10 \\ 20 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ .

So $x = 1$ , $y = 2$ . Markers reward the determinant, the inverse, and the matrix-product solution.

AQA 20224 marksDescribe fully the single geometric transformation represented by the matrix

\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

, and state its determinant and what that determinant means.

Show worked answer →

Read the columns as the images of the basis vectors. The first column is the image of $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , which maps to $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ ; the second column is the image of $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , which maps to $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .

Swapping the coordinates of every point is a reflection in the line $y = x$ .

The determinant is $(0)(0) - (1)(1) = -1$ . The magnitude $1$ means areas are preserved, and the negative sign means orientation is reversed, which is consistent with a reflection.

Markers reward identifying the reflection in $y = x$ , the determinant $-1$ , and linking the sign to reversed orientation.

Related dot points

Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)