Determinant and inverse of a 3×3 matrix

Matrices

A-level Further Maths (9FM0) · Core Pure

ORIGINAL

Determinant, minors, cofactors, transpose, divide — the full 3×3 inverse with a check at the end.

Board and spec code confirmed against Edexcel 9FM0 · registry checked 2026-07-17How this checking works

Question

\mathbf{M} = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 1 & 0 & 2 \end{pmatrix}

. Find

\det \mathbf{M}

and hence find

\mathbf{M}^{-1}

Every step worked, with the reasoning.

1
$\det \mathbf{M} = 1\begin{vmatrix} 1 & 3 \\ 0 & 2 \end{vmatrix} - 2\begin{vmatrix} 0 & 3 \\ 1 & 2 \end{vmatrix} + 1\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix}$
Expand along the top row with the sign pattern $+\ -\ +$ .
2
$\det \mathbf{M} = 1(2) - 2(-3) + 1(-1) = 7$
$\det \mathbf{M} \ne 0$ , so $\mathbf{M}$ is non-singular and $\mathbf{M}^{-1}$ exists.
3
Matrix of minors: $\begin{pmatrix} 2 & -3 & -1 \\ 4 & 1 & -2 \\ 5 & 3 & 1 \end{pmatrix}$
Each entry is the $2 \times 2$ determinant left after deleting that entry's row and column.
4
Matrix of cofactors: $\begin{pmatrix} 2 & 3 & -1 \\ -4 & 1 & 2 \\ 5 & -3 & 1 \end{pmatrix}$
Apply the sign pattern $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ to the minors.
5
$\mathbf{M}^{-1} = \dfrac{1}{7}\begin{pmatrix} 2 & -4 & 5 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$
Transpose the cofactor matrix (rows become columns) to get the adjugate, then divide by $\det \mathbf{M} = 7$ .
6
Check row 1 of $\mathbf{M}\mathbf{M}^{-1}$ : $\tfrac{1}{7}(2 + 6 - 1,\ -4 + 2 + 2,\ 5 - 6 + 1) = (1,\ 0,\ 0)$
Multiplying back to the identity catches a sign slip before it costs the accuracy mark.

Answer: $\det \mathbf{M} = 7$ , $\ \mathbf{M}^{-1} = \dfrac{1}{7}\begin{pmatrix} 2 & -4 & 5 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$