Find the determinant of pmatrix 3 & 5 \\ 2 & 4 pmatrix. [1 mark]

What does a final row 0\,0\,0 0 tell you about a 3× 3 system? [1 mark]

SQA AH style: 2x2 inverse-style practice: Find the inverse of A = pmatrix 2 & 1 \\ 3 & 4 pmatrix and hence solve 2x + y = 5, 3x + 4y = 10.

Determinant A = (2)(4) - (1)(3) = 5 (1 mark). Inverse A^-1 = 15pmatrix 4 & -1 \\ -3 & 2 pmatrix (1 mark). Solution pmatrix x \\ y pmatrix = A^-1pmatrix 5 \\ 10 pmatrix = 15pmatrix 20 - 10 \\ -15 + 20 pmatrix = 15pmatrix 10 \\ 5 pmatrix (1 mark). So x = 2, y = 1 (1 mark). Markers reward the determinant, the inverse with the swap-and-negate pattern, the matrix product, and the solution.

ScotlandMathsSyllabus dot point

How do you compute with matrices and use them to solve a system of linear equations?

Add, subtract and multiply matrices, find the determinant and inverse of 2x2 and 3x3 matrices, and solve systems of linear equations using the inverse matrix and Gaussian elimination, identifying unique, no, and infinitely many solutions.

A focused answer to the SQA Advanced Higher Mathematics matrices and systems of equations content, covering matrix addition, subtraction and multiplication, the determinant and inverse of 2x2 and 3x3 matrices, solving systems by the inverse matrix and by Gaussian elimination, and recognising unique, no, and infinitely many solutions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

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

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What this dot point is asking
Matrix arithmetic
Determinant and inverse
Solving systems of equations
The three outcomes
Try this

What this dot point is asking

The SQA wants you to compute with matrices, find determinants and inverses of $2\times 2$ and $3\times 3$ matrices, and use both the inverse-matrix method and Gaussian elimination to solve systems of linear equations, recognising when a system has a unique solution, no solution, or infinitely many.

Matrix arithmetic

Addition and subtraction act entry by entry on matrices of the same size. Multiplication combines the rows of the first matrix with the columns of the second, so the number of columns of the first must equal the number of rows of the second. Order matters: in general $AB \ne BA$ .

Multiplying two

2 \times 2

matrices

Multiply $\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 4 & 1 \\ 2 & 5 \end{pmatrix}$ .

Step 1: Multiply row 1 of the first matrix into each column of the second

Each entry in the top row of the product is the dot product of row 1 with the corresponding column:

\text{Top-left: } 1(4) + 2(2) = 8;\quad \text{Top-right: } 1(1) + 2(5) = 11.

Step 2: Multiply row 2 of the first matrix into each column of the second

Repeat with the bottom row. Row 2 starts with $0$ , which immediately kills the first column of the second matrix:

\text{Bottom-left: } 0(4) + 3(2) = 6;\quad \text{Bottom-right: } 0(1) + 3(5) = 15.

Step 3: Assemble the product matrix

Place the four computed entries in the corresponding positions:

\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 4 & 1 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 8 & 11 \\ 6 & 15 \end{pmatrix}.

Final answer: the product is $\begin{pmatrix} 8 & 11 \\ 6 & 15 \end{pmatrix}$ .

Determinant and inverse

The determinant decides whether a matrix is invertible: a matrix with determinant zero is singular and has no inverse. For a $2\times 2$ matrix the inverse follows a swap-and-negate pattern; for a $3\times 3$ matrix the determinant is expanded along a row using cofactors.

Solving systems of equations

Two methods solve a linear system. The inverse-matrix method writes the system as $A\mathbf{x} = \mathbf{b}$ and computes $\mathbf{x} = A^{-1}\mathbf{b}$ , which works only when $\det A \ne 0$ . Gaussian elimination is more general: form the augmented matrix and use row operations to reach upper-triangular form, then back-substitute.

Solving a

2\times 2

system by Gaussian elimination

Solve by Gaussian elimination: $x + 2y = 5$ , $3x - y = 1$ .

Step 1: Form the augmented matrix

Write the coefficients and constants as a matrix. The vertical line separates the left-hand side from the right-hand side of each equation:

\left(\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & -1 & 1 \end{array}\right).

Step 2: Eliminate $x$ from row 2

Subtract three times row 1 from row 2. This creates a zero in the bottom-left, leaving a single equation in $y$ :

R_2 - 3R_1:\quad \left(\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -7 & -14 \end{array}\right).

Step 3: Back-substitute to find $x$ and $y$

Read the bottom row as $-7y = -14$ , giving $y = 2$ . Substitute into the top equation:

x + 2(2) = 5, \quad x = 1.

Final answer: $x = 1$ , $y = 2$ .

The three outcomes

After elimination, read off the type of solution from the final rows. A full set of leading entries gives a unique solution. A row reading $0 =$ (non-zero), such as $0\,0\,0 \mid 4$ , is a contradiction, so there is no solution. A row of all zeros, $0\,0\,0 \mid 0$ , removes one equation, leaving a free variable and infinitely many solutions, which you express in terms of a parameter.

Geometrically a $3\times 3$ system is the question of how three planes sit in space. A unique solution is a single common point; no solution means the planes have no point in common, as with three planes forming a triangular prism; and infinitely many solutions means the three planes share a whole line. This is exactly the connection to the vectors topic, where a plane is described by a point and a normal, so the algebra of elimination and the geometry of intersecting planes are two views of one problem. Recognising the geometric meaning is often the quickest way to interpret an unexpected row of zeros.

Try this

Q1. Find the determinant of $\begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}$ . [1 mark]

Cue. $3(4) - 5(2) = 12 - 10 = 2$ .

Q2. What does a final row $0\,0\,0 \mid 0$ tell you about a $3\times 3$ system? [1 mark]

Cue. One equation is redundant, so there are infinitely many solutions (a parameter is needed).

Exam-style practice questions

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

AH style: 2x2 inverse4 marksFind the inverse of

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

and hence solve

2x + y = 5

3x + 4y = 10

Show worked answer →

Determinant $\det A = (2)(4) - (1)(3) = 5$ (1 mark).

Inverse $A^{-1} = \dfrac{1}{5}\begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}$ (1 mark).

Solution $\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 20 - 10 \\ -15 + 20 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 10 \\ 5 \end{pmatrix}$ (1 mark).

So $x = 2$ , $y = 1$ (1 mark). Markers reward the determinant, the inverse with the swap-and-negate pattern, the matrix product, and the solution.

AH style: Gaussian elimination5 marksUse Gaussian elimination to solve

x + y + z = 6

2x - y + z = 3

x + 2y - z = 2

Show worked answer →

Form the augmented matrix and eliminate $x$ from rows 2 and 3 using row 1: $R_2 - 2R_1$ gives $-3y - z = -9$ ; $R_3 - R_1$ gives $y - 2z = -4$ (2 marks).

Eliminate $y$ : from $-3y - z = -9$ and $y - 2z = -4$ , use $3(y - 2z) + (-3y - z)$ to get $-7z = -21$ , so $z = 3$ (1 mark).

Back-substitute: $y - 2(3) = -4$ gives $y = 2$ ; then $x + 2 + 3 = 6$ gives $x = 1$ (1 mark).

Solution $x = 1$ , $y = 2$ , $z = 3$ (1 mark). Markers reward the elimination of $x$ , the elimination of $y$ , the value of $z$ , and the back-substitution.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)