AS & A Level · Practice questions · Determinants and inverses

Find the determinant, then divide.

Six original Cambridge-style questions on determinants and inverses: 2 by 2 and 3 by 3 determinants, singular matrices, the adjugate method, \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) and an inverse from a matrix identity.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Determinant first, then build, then check.

Determinant: \(ad - bc\), or expand a 3 by 3 along row 1 with signs \(+\ -\ +\). Zero means singular: stop.
2 by 2 inverse: swap the leading diagonal, negate the other two, divide by the determinant.
3 by 3 inverse: minors, then cofactors (sign pattern), then transpose to the adjugate, then divide by the determinant.
Products: \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) and \(\det(\mathbf{AB}) = \det\mathbf{A}\det\mathbf{B}\).
Check: multiply by the original matrix and look for \(\mathbf{I}\). Use your calculator to confirm, but show the working.

Recall

[2 marks]

Find the inverse of \(\begin{pmatrix}6&-2\\5&-1\end{pmatrix}\).

Determinant: \(6(-1) - (-2)(5) = -6 + 10 = 4\). ✓
Inverse: \(\dfrac{1}{4}\begin{pmatrix}-1&2\\-5&6\end{pmatrix}\), that is \(\begin{pmatrix}-\frac{1}{4}&\frac{1}{2}\\-\frac{5}{4}&\frac{3}{2}\end{pmatrix}\). ✓

Calculation

[4 marks]

The matrix \(\mathbf{M}\) is \(\begin{pmatrix}k&3\\2&k + 5\end{pmatrix}\).

(a) Find the values of \(k\) for which \(\mathbf{M}\) is singular.

(b) Find \(\mathbf{M}^{-1}\) when \(k = 2\).

(a) \(\det\mathbf{M} = k(k + 5) - 6 = k^{2} + 5k - 6 = (k + 6)(k - 1)\). ✓
Singular when \(\det\mathbf{M} = 0\): \(k = 1\) or \(k = -6\). ✓
(b) \(k = 2\): \(\mathbf{M} = \begin{pmatrix}2&3\\2&7\end{pmatrix}\), with determinant \(14 - 6 = 8\). ✓
\(\mathbf{M}^{-1} = \dfrac{1}{8}\begin{pmatrix}7&-3\\-2&2\end{pmatrix}\). ✓

Analysis

[4 marks]

Show that the matrix \(\begin{pmatrix}2&k&1\\-k&1&0\\1&0&1\end{pmatrix}\) is non-singular for all real values of \(k\).

Expand along row 1: \(2\begin{vmatrix}1&0\\0&1\end{vmatrix} - k\begin{vmatrix}-k&0\\1&1\end{vmatrix} + 1\begin{vmatrix}-k&1\\1&0\end{vmatrix}\). ✓
\(= 2(1) - k(-k - 0) + (0 - 1) = 2 + k^{2} - 1\). ✓
\(= k^{2} + 1\). ✓
\(k^{2} \ge 0\), so the determinant is at least 1 and never zero: the matrix is non-singular for every real \(k\). ✓

Calculation

[6 marks]

\(\mathbf{Q} = \begin{pmatrix}1&2&0\\0&1&1\\2&0&1\end{pmatrix}\).

(a) Find \(\det\mathbf{Q}\).

(b) Find \(\mathbf{Q}^{-1}\), showing the matrix of cofactors.

(a) \(1(1 - 0) - 2(0 - 2) + 0 = 1 + 4 = 5\). ✓
(b) Minors: \(\begin{pmatrix}1&-2&-2\\2&1&-4\\2&1&1\end{pmatrix}\). ✓
Cofactors (apply \(+\ -\ +\,/\,-\ +\ -\,/\,+\ -\ +\)): \(\begin{pmatrix}1&2&-2\\-2&1&4\\2&-1&1\end{pmatrix}\). ✓
Adjugate (transpose): \(\begin{pmatrix}1&-2&2\\2&1&-1\\-2&4&1\end{pmatrix}\), so \(\mathbf{Q}^{-1} = \dfrac{1}{5}\begin{pmatrix}1&-2&2\\2&1&-1\\-2&4&1\end{pmatrix}\). ✓
(c) Row 1 of \(\mathbf{Q}\), \((1,\ 2,\ 0)\), times the adjugate's columns: \(1 + 4 = 5\), \(-2 + 2 = 0\), \(2 - 2 = 0\). ✓
Row 2, \((0,\ 1,\ 1)\): \(2 - 2 = 0\), \(1 + 4 = 5\), \(-1 + 1 = 0\). Row 3, \((2,\ 0,\ 1)\): \(2 - 2 = 0\), \(-4 + 4 = 0\), \(4 + 1 = 5\). ✓
So \(\mathbf{Q}\operatorname{adj}\mathbf{Q} = 5\mathbf{I}\), and dividing by 5 gives \(\mathbf{Q}\mathbf{Q}^{-1} = \mathbf{I}\). ✓

Proof

[5 marks]

(a) \(\mathbf{A}\) and \(\mathbf{B}\) are non-singular square matrices of the same size. Prove that \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\).

(b) Given that \(\mathbf{A}^{-1} = \begin{pmatrix}1&1\\0&2\end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix}3&1\\2&1\end{pmatrix}\), find \((\mathbf{AB})^{-1}\), and check your answer using determinants.

(a) \((\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) = \mathbf{A}(\mathbf{B}\mathbf{B}^{-1})\mathbf{A}^{-1} = \mathbf{A}\mathbf{I}\mathbf{A}^{-1} = \mathbf{A}\mathbf{A}^{-1} = \mathbf{I}\). ✓
A matrix that multiplies \(\mathbf{AB}\) to give \(\mathbf{I}\) is its inverse, so \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). ✓
(b) \(\det\mathbf{B} = 3 - 2 = 1\), so \(\mathbf{B}^{-1} = \begin{pmatrix}1&-1\\-2&3\end{pmatrix}\). ✓
\((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{pmatrix}1&-1\\-2&3\end{pmatrix}\begin{pmatrix}1&1\\0&2\end{pmatrix} = \begin{pmatrix}1&-1\\-2&4\end{pmatrix}\). ✓
Check: \(\det\mathbf{B}^{-1}\det\mathbf{A}^{-1} = 1 \times 2 = 2\), and \(\begin{vmatrix}1&-1\\-2&4\end{vmatrix} = 4 - 2 = 2\). ✓

Analysis

[5 marks]

\(\mathbf{A} = \begin{pmatrix}2&1\\-1&0\end{pmatrix}\).

(a) Show that \(\mathbf{A}^{2} = 2\mathbf{A} - \mathbf{I}\).

(b) Hence show that \(\mathbf{A}^{-1} = 2\mathbf{I} - \mathbf{A}\), and write down \(\mathbf{A}^{-1}\).

(a) \(\mathbf{A}^{2} = \begin{pmatrix}2&1\\-1&0\end{pmatrix}\begin{pmatrix}2&1\\-1&0\end{pmatrix} = \begin{pmatrix}4 - 1 & 2 + 0\\-2 + 0 & -1 + 0\end{pmatrix} = \begin{pmatrix}3&2\\-2&-1\end{pmatrix}\). ✓
\(2\mathbf{A} - \mathbf{I} = \begin{pmatrix}4&2\\-2&0\end{pmatrix} - \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}3&2\\-2&-1\end{pmatrix}\), the same. ✓
(b) Rearrange: \(2\mathbf{A} - \mathbf{A}^{2} = \mathbf{I}\), so \(\mathbf{A}(2\mathbf{I} - \mathbf{A}) = \mathbf{I}\). Hence \(\mathbf{A}^{-1} = 2\mathbf{I} - \mathbf{A}\). ✓
\(\mathbf{A}^{-1} = \begin{pmatrix}2&0\\0&2\end{pmatrix} - \begin{pmatrix}2&1\\-1&0\end{pmatrix} = \begin{pmatrix}0&-1\\1&2\end{pmatrix}\). ✓
(c) \(\det\mathbf{A} = 0 - (-1) = 1\), so \(\mathbf{A}^{-1} = \begin{pmatrix}0&-1\\1&2\end{pmatrix}\), which agrees. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.