What is solving a system by the inverse?

A system of n equations in n unknowns is Ax = b. If A is invertible, the unique solution is

Write the inverse of pmatrix 2 & 0 \\ 1 & 3 pmatrix. [2 marks]

A 33 system has A = 0 and reduces to a row (0\ 0\ 0 4). What can you conclude? [1 mark]

SEAB Original-style practice: Find the inverse of A = pmatrix 4 & 3 \\ 2 & 2 pmatrix and hence solve cases 4x + 3y = 1 \\ 2x + 2y = 0 cases.

Determinant: A = 4(2) - 3(2) = 8 - 6 = 2 ≠ 0, so A is invertible. For a 22 matrix, A^-1 = 1Apmatrix d & -b \\ -c & a pmatrix = 12pmatrix 2 & -3 \\ -2 & 4 pmatrix = pmatrix 1 & -32 \\ -1 & 2 pmatrix. Write the system as Apmatrix x \\ y pmatrix = pmatrix 1 \\ 0 pmatrix, so pmatrix x \\ y pmatrix = A^-1pmatrix 1 \\ 0 pmatrix = pmatrix 1 \\ -1 pmatrix. Hence x = 1, y = -1. Markers reward the non-zero determinant, the swap-and-negate inverse formula, expressing the system as Ax = b, and the solution x = 1, y = -1.

SingaporeFurther MathsSyllabus dot point

How do we invert a matrix and use it to solve a system of linear equations?

Find the inverse of a non-singular matrix and use matrices to solve systems of linear equations, recognising consistent, inconsistent and dependent cases

A focused answer to the H2 Further Mathematics outcome on inverse matrices and linear systems. The 2x2 inverse formula, the adjugate method and row reduction for 3x3, solving systems by the inverse, and classifying consistent, inconsistent and dependent systems.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find the inverse of a non-singular square matrix and use it to solve a system of linear equations written in the form $\mathbf{Ax} = \mathbf{b}$ . You must also classify a system as having a unique solution, no solution (inconsistent), or infinitely many solutions (dependent), and interpret each case.

The answer

When an inverse exists

A square matrix $\mathbf{A}$ has an inverse $\mathbf{A}^{-1}$ (with $\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$ ) if and only if $\det\mathbf{A} \neq 0$ . If $\det\mathbf{A} = 0$ the matrix is singular and has no inverse.

The 2x2 inverse

For $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $\det\mathbf{A} = ad - bc \neq 0$ ,

\mathbf{A}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

Swap the leading diagonal, negate the off-diagonal, and divide by the determinant.

The 3x3 inverse

Two reliable methods:

Adjugate method: $\mathbf{A}^{-1} = \dfrac{1}{\det\mathbf{A}}\operatorname{adj}\mathbf{A}$ , where the adjugate is the transpose of the cofactor matrix.
Row reduction: augment $[\mathbf{A} \mid \mathbf{I}]$ and apply row operations until the left block becomes $\mathbf{I}$ ; the right block is then $\mathbf{A}^{-1}$ .

In practice the graphing calculator computes a numerical inverse, but the method must be shown when a question asks for it.

Solving a system by the inverse

A system of $n$ equations in $n$ unknowns is $\mathbf{Ax} = \mathbf{b}$ . If $\mathbf{A}$ is invertible, the unique solution is

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

Classifying systems

When $\det\mathbf{A} \neq 0$ there is a unique solution. When $\det\mathbf{A} = 0$ the system is either:

inconsistent: the equations contradict (parallel planes with no common point), giving no solution; or
dependent: the equations are not independent, giving infinitely many solutions described by one or more parameters (planes meeting in a line, or all the same plane).

Row reduction reveals which case holds: a row $\begin{pmatrix} 0 & 0 & 0 \mid c \end{pmatrix}$ with $c \neq 0$ is inconsistent, while $\begin{pmatrix} 0 & 0 & 0 \mid 0 \end{pmatrix}$ signals a free parameter.

Worked example

Solve the system $\begin{cases} x + 2y = 5 \\ 3x + 4y = 6 \end{cases}$ using the inverse matrix.

Step 1: Write in matrix form

\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}.

Step 2: Compute the determinant

\det\mathbf{A} = (1)(4) - (2)(3) = 4 - 6 = -2 \neq 0,

so the inverse exists and the solution is unique.

Step 3: Form the inverse

\mathbf{A}^{-1} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \tfrac{3}{2} & -\tfrac{1}{2} \end{pmatrix}.

Step 4: Multiply by b

\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \tfrac{3}{2} & -\tfrac{1}{2} \end{pmatrix}\begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -10 + 6 \\ \tfrac{15}{2} - 3 \end{pmatrix} = \begin{pmatrix} -4 \\ \tfrac{9}{2} \end{pmatrix}.

Step 5: State the solution

$x = -4$ , $y = \tfrac{9}{2}$ . Substituting back into the original equations confirms both hold.

Examples in context

Example 1. Balancing a network. Currents in a circuit obeying Kirchhoff's laws, or flows in a transport network, form a linear system $\mathbf{Ax} = \mathbf{b}$ . A non-singular $\mathbf{A}$ gives a unique set of currents; a singular one signals a redundancy or an impossible demand.

Example 2. Three planes in space. Three linear equations in $x, y, z$ are three planes. They meet in a single point (unique solution), in a common line or plane (dependent, infinitely many), or in no common point (inconsistent), exactly mirroring the algebraic classification.

Try this

Q1. State the condition for a square matrix to be invertible. [1 mark]

Cue. Its determinant is non-zero ( $\det\mathbf{A} \neq 0$ ).

Q2. Write the inverse of $\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$ . [2 marks]

Cue. $\det = 6$ , so $\dfrac{1}{6}\begin{pmatrix} 3 & 0 \\ -1 & 2 \end{pmatrix}$ .

Q3. A $3\times3$ system has $\det\mathbf{A} = 0$ and reduces to a row $(0\ 0\ 0 \mid 4)$ . What can you conclude? [1 mark]

Cue. The system is inconsistent and has no solution.

Exam-style practice questions

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

Original5 marksFind the inverse of

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

and hence solve

\begin{cases} 4x + 3y = 1 \\ 2x + 2y = 0 \end{cases}

Show worked answer →

Determinant: $\det\mathbf{A} = 4(2) - 3(2) = 8 - 6 = 2 \neq 0$ , so $\mathbf{A}$ is invertible.

For a $2\times2$ matrix, $\mathbf{A}^{-1} = \dfrac{1}{\det\mathbf{A}}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} 2 & -3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & -\tfrac{3}{2} \\ -1 & 2 \end{pmatrix}$ .

Write the system as $\mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , so $\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{A}^{-1}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

Hence $x = 1$ , $y = -1$ .

Markers reward the non-zero determinant, the swap-and-negate inverse formula, expressing the system as $\mathbf{Ax} = \mathbf{b}$ , and the solution $x = 1$ , $y = -1$ .

Original6 marksThe system

\begin{cases} x + y + z = 2 \\ 2x + 3y + z = 3 \\ 3x + 4y + 2z = k \end{cases}

is to be solved. Determine the value of

k

for which the system is consistent, and describe the solution set in that case.

Show worked answer →

Add the first two equations cleverly or use elimination. Note that (row 1) + (row 2) gives $3x + 4y + 2z = 5$ . The third equation says $3x + 4y + 2z = k$ .

For consistency the two expressions for $3x + 4y + 2z$ must agree, so $k = 5$ .

When $k = 5$ the third equation is the sum of the first two, so it carries no new information; the system reduces to two independent equations in three unknowns. There is therefore a one-parameter family of solutions (a line of solutions), not a unique point.

Let $z = t$ . From row 1, $x + y = 2 - t$ ; from row 2, $2x + 3y = 3 - t$ . Solving: $y = (3 - t) - 2(2 - t) = t - 1$ , and $x = (2 - t) - (t - 1) = 3 - 2t$ . So $(x, y, z) = (3 - 2t,\ t - 1,\ t)$ .

Markers reward spotting the dependence, the condition $k = 5$ for consistency, recognising a one-parameter (line) solution set, and a correct parametric solution.