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

Given A = 3 and B = -2 for 3 × 3 matrices, find (AB^-1). [2 marks]

OCR OCR 2018-style practice: Find the inverse of A = pmatrix 4 & 3 \\ 5 & 4 pmatrix, and verify your answer by computing AA^-1.

Determinant (M1): A = 4(4) - 3(5) = 16 - 15 = 1. Inverse using the 2 × 2 rule (swap leading diagonal, negate the other diagonal, divide by the determinant) (M1, A1): A^-1 = 11pmatrix 4 & -3 \\ -5 & 4 pmatrix = pmatrix 4 & -3 \\ -5 & 4 pmatrix. Verify (A1): AA^-1 = pmatrix 4 & 3 \\ 5 & 4 pmatrixpmatrix 4 & -3 \\ -5 & 4 pmatrix = pmatrix 16 - 15 & -12 + 12 \\ 20 - 20 & -15 + 16 pmatrix = pmatrix 1 & 0 \\ 0 & 1 pmatrix = I. Markers reward the determinant, the correct rearrangement of entries, and a verification that returns the identity.

OCR OCR 2022-style practice: The matrix B = pmatrix 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 1 & 1 pmatrix. Find B and hence find B^-1.

Determinant by expanding along the first row (M1, A1): B = 1(1 · 1 - 0 · 1) - 0 + 2(2 · 1 - 1 · 0) = 1(1) + 2(2) = 5. Since B = 5 0, the inverse exists. Form the matrix of cofactors, transpose to the adjugate, and divide by the determinant (M1 for cofactors, A1 for the adjugate, M1, A1 for the final inverse): B^-1 = 15pmatrix 1 & 2 & -2 \\ -2 & 1 & 4 \\ 2 & -1 & 1 pmatrix. Markers reward the determinant, a correct cofactor matrix, the transpose to the adjugate, and dividing by the determinant. A calculator may be used to check, but the method must be shown.

EnglandFurther MathsSyllabus dot point

How do you find the inverse of a 2x2 and 3x3 matrix, and when does an inverse exist?

The inverse of a 2x2 matrix, the existence condition (non-zero determinant), the inverse of a 3x3 matrix via the adjugate or row reduction, and the inverse of a product.

A focused answer to the OCR A-Level Further Mathematics A content on inverse matrices, covering the formula for the inverse of a 2x2 matrix, the condition for an inverse to exist, finding the inverse of a 3x3 matrix using the adjugate (matrix of cofactors) or row reduction, and the rule for the inverse of a product of matrices.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

A square matrix $\mathbf{A}$ has an inverse $\mathbf{A}^{-1}$ exactly when $\det \mathbf{A} \ne 0$ ; such a matrix is non-singular. For a $2 \times 2$ matrix, $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ : swap the leading diagonal, negate the other diagonal, and divide by the determinant. For a $3 \times 3$ matrix, $\mathbf{A}^{-1} = \dfrac{1}{\det \mathbf{A}}\,\text{adj}(\mathbf{A})$ , where the adjugate is the transpose of the matrix of cofactors. The inverse satisfies $\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$ , and the inverse of a product reverses the order: $(\mathbf{A}\mathbf{B})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$ .

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What this dot point is asking
The inverse of a 2x2 matrix
When an inverse exists
The inverse of a 3x3 matrix
The inverse of a product
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What this dot point is asking

OCR wants you to find the inverse of a $2 \times 2$ matrix from the standard formula, state the condition under which an inverse exists (a non-zero determinant), find the inverse of a $3 \times 3$ matrix using the adjugate (the transposed matrix of cofactors) or by row reduction, and use the rule for the inverse of a product. The inverse is the gateway to solving matrix equations and systems of linear equations.

The inverse of a 2x2 matrix

The inverse undoes the matrix: $\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}$ . For a $2 \times 2$ matrix there is a quick formula, valid whenever the determinant is non-zero.

The mnemonic is "swap, negate, divide": swap the entries on the leading diagonal, negate the entries on the other diagonal, then divide every entry by the determinant.

When an inverse exists

A square matrix is invertible (non-singular) precisely when $\det \mathbf{A} \ne 0$ . If the determinant is zero the transformation collapses space onto a lower dimension, information is lost, and no inverse can recover the original; the matrix is singular. This single condition governs every inverse question and every "for what values of $k$ is this invertible" problem.

The inverse of a 3x3 matrix

For a $3 \times 3$ matrix the standard route is the adjugate method. Build the matrix of cofactors (each entry is the signed minor obtained by deleting that entry's row and column), transpose it to get the adjugate, then divide by the determinant.

In the exam a calculator can produce the inverse directly, but OCR expects you to show the method (determinant, cofactors, adjugate) when the question says "find" or "show", and you should always verify with $\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}$ .

The inverse of a product

For invertible matrices, the inverse of a product reverses the order. This matters when you unpick a chain of transformations.

The inverse is the tool you reach for whenever a matrix equation must be solved, and it leads directly into solving systems of linear equations.

Try this

Q1. Find the inverse of $\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$ . [3 marks]

Cue. $\det = 3 - 2 = 1$ , so the inverse is $\begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix}$ .

Q2. Given $\det \mathbf{A} = 3$ and $\det \mathbf{B} = -2$ for $3 \times 3$ matrices, find $\det(\mathbf{A}\mathbf{B}^{-1})$ . [2 marks]

Cue. $\det(\mathbf{A}\mathbf{B}^{-1}) = \det \mathbf{A} \times \dfrac{1}{\det \mathbf{B}} = 3 \times \left(-\tfrac{1}{2}\right) = -\tfrac{3}{2}$ .

Exam-style practice questions

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

OCR 20184 marksFind the inverse of

\mathbf{A} = \begin{pmatrix} 4 & 3 \\ 5 & 4 \end{pmatrix}

, and verify your answer by computing

\mathbf{A}\mathbf{A}^{-1}

Show worked answer →

Determinant (M1): $\det \mathbf{A} = 4(4) - 3(5) = 16 - 15 = 1$ .

Inverse using the $2 \times 2$ rule (swap leading diagonal, negate the other diagonal, divide by the determinant) (M1, A1): $\mathbf{A}^{-1} = \dfrac{1}{1}\begin{pmatrix} 4 & -3 \\ -5 & 4 \end{pmatrix} = \begin{pmatrix} 4 & -3 \\ -5 & 4 \end{pmatrix}$ .

Verify (A1): $\mathbf{A}\mathbf{A}^{-1} = \begin{pmatrix} 4 & 3 \\ 5 & 4 \end{pmatrix}\begin{pmatrix} 4 & -3 \\ -5 & 4 \end{pmatrix} = \begin{pmatrix} 16 - 15 & -12 + 12 \\ 20 - 20 & -15 + 16 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I}$ .

Markers reward the determinant, the correct rearrangement of entries, and a verification that returns the identity.

OCR 20226 marksThe matrix

\mathbf{B} = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}

. Find

\det \mathbf{B}

and hence find

\mathbf{B}^{-1}

Show worked answer →

Determinant by expanding along the first row (M1, A1): $\det \mathbf{B} = 1(1 \cdot 1 - 0 \cdot 1) - 0 + 2(2 \cdot 1 - 1 \cdot 0) = 1(1) + 2(2) = 5$ .

Since $\det \mathbf{B} = 5 \ne 0$ , the inverse exists. Form the matrix of cofactors, transpose to the adjugate, and divide by the determinant (M1 for cofactors, A1 for the adjugate, M1, A1 for the final inverse):

$\mathbf{B}^{-1} = \dfrac{1}{5}\begin{pmatrix} 1 & 2 & -2 \\ -2 & 1 & 4 \\ 2 & -1 & 1 \end{pmatrix}$ .

Markers reward the determinant, a correct cofactor matrix, the transpose to the adjugate, and dividing by the determinant. A calculator may be used to check, but the method must be shown.

Related dot points

Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)