AS June 2025 Paper 1 Q14
14
(a) The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by\[\mathbf{A} = \begin{bmatrix} 4 & -3 \\ -1 & 1 \end{bmatrix} \qquad \text{and} \qquad \mathbf{B} = \begin{bmatrix} 5 & 4 \\ -3 & -2 \end{bmatrix}\]
(i) Find the matrices \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1}\) [2 marks]
(ii) Hence verify that \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) [2 marks]
(b) Given that \((\mathbf{CD})^{-1} = \mathbf{D}^{-1}\mathbf{C}^{-1}\) is true for all non-singular square matrices \(\mathbf{C}\) and \(\mathbf{D}\), prove by induction that\[(\mathbf{M}^{-1})^n = (\mathbf{M}^n)^{-1}\]
is true for all \(n \in \mathbb{N}\), where \(\mathbf{M}\) is a non-singular square matrix. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(\begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}\) | B1 | 1.1b |
| Obtains \(\begin{bmatrix} -1 & -2 \\ 1.5 & 2.5 \end{bmatrix}\) or \(\dfrac{1}{2}\begin{bmatrix} -2 & -4 \\ 3 & 5 \end{bmatrix}\) or \(-\dfrac{1}{2}\begin{bmatrix} 2 & 4 \\ -3 & -5 \end{bmatrix}\) | B1 | 1.1b |
| (2) | ||
| (ii) Obtains at least three correct elements of \(\mathbf{AB}\) May be unsimplified. | M1 | 1.1a |
| Correctly verifies that \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) Condone missing conclusion. | R1 | 2.2a |
| (2) |
Typical solution
(i)
\[|\mathbf{A}| = 4 - 3 = 1\]\[\mathbf{A}^{-1} = \begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}\]\[|\mathbf{B}| = -10 + 12 = 2\]\[\mathbf{B}^{-1} = \frac{1}{2}\begin{bmatrix} -2 & -4 \\ 3 & 5 \end{bmatrix}\](ii)
\[\begin{aligned}\mathbf{AB} &= \begin{bmatrix} 4 & -3 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 5 & 4 \\ -3 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 29 & 22 \\ -8 & -6 \end{bmatrix}\end{aligned}\]\[(\mathbf{AB})^{-1} = \frac{1}{2}\begin{bmatrix} -6 & -22 \\ 8 & 29 \end{bmatrix}\]\[\begin{aligned}\mathbf{B}^{-1}\mathbf{A}^{-1} &= \frac{1}{2}\begin{bmatrix} -2 & -4 \\ 3 & 5 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix} -6 & -22 \\ 8 & 29 \end{bmatrix}\end{aligned}\]so \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)
| Scheme | Marks | AO |
|---|---|---|
| Writes \((\mathbf{M}^{-1})^1 = \mathbf{M}^{-1}\) and \((\mathbf{M}^1)^{-1} = \mathbf{M}^{-1}\) | B1 | 2.2a |
| Assumes \((\mathbf{M}^{-1})^k = (\mathbf{M}^k)^{-1}\) and considers \((\mathbf{M}^{-1})^k\,\mathbf{M}^{-1}\) or \(\mathbf{M}^{-1}(\mathbf{M}^{-1})^k\) | M1 | 2.4 |
| Completes working to show \((\mathbf{M}^k)^{-1}\mathbf{M}^{-1}\) or \(\mathbf{M}^{-1}(\mathbf{M}^{-1})^k\) is equivalent to \((\mathbf{M}^{k+1})^{-1}\) | A1 | 2.2a |
| Completes a reasoned argument by stating that the rule is true for \(n = 1\) and if the rule is true for \(n = k\) then it is also true for \(n = k + 1\) and concludes that (by induction) \((\mathbf{M}^{-1})^n = (\mathbf{M}^n)^{-1}\) is true for all \(n \in \mathbf{N}\) This mark is dependent on all previous marks. The algebra must use an alternative letter to \(n\) Condone reference to ‘rule’/‘statement’/‘it’ in the concluding statement. | R1 | 2.1 |
| (4) | ||
| (8 marks) |
Typical solution
When \(n = 1\):
\[(\mathbf{M}^{-1})^1 = \mathbf{M}^{-1} \text{ and } (\mathbf{M}^1)^{-1} = \mathbf{M}^{-1}\]so, the rule is true for \(n = 1\)
Assume the rule is true for \(n = k\)
\[\begin{aligned}(\mathbf{M}^{-1})^k &= (\mathbf{M}^k)^{-1} \\ (\mathbf{M}^{-1})^k\,\mathbf{M}^{-1} &= (\mathbf{M}^k)^{-1}\,\mathbf{M}^{-1} \\ (\mathbf{M}^{-1})^{k+1} &= (\mathbf{M}\mathbf{M}^k)^{-1} \\ &= (\mathbf{M}^{k+1})^{-1}\end{aligned}\]so, the rule is also true for \(n = k + 1\)
So, by induction,
\[(\mathbf{M}^{-1})^n = (\mathbf{M}^n)^{-1}\]is true for all \(n \in \mathbf{N}\)