AS June 2022 Paper 1 Q11
11 Prove by induction that, for all integers \(n \geqslant 1\),
\[(\mathbf{ABA}^{-1})^{n} = \mathbf{AB}^{n}\mathbf{A}^{-1}\]where \(\mathbf{A}\) and \(\mathbf{B}\) are square matrices of equal dimensions, and \(\mathbf{A}\) is non-singular. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows that \((\mathbf{ABA}^{-1})^{n} = \mathbf{AB}^{n}\mathbf{A}^{-1}\) is true for \(n = 1\) | B1 | 2.1 |
| Assumes \((\mathbf{ABA}^{-1})^{k} = \mathbf{AB}^{k}\mathbf{A}^{-1}\) and multiplies by \(\mathbf{ABA}^{-1}\) | M1 | 2.4 |
| Completes rigorous working to show \((\mathbf{ABA}^{-1})^{k+1} = \mathbf{AB}^{k+1}\mathbf{A}^{-1}\) Condone \(\mathbf{A}^{-1}\mathbf{A}\) removed without reference to \(\mathbf{I}\) | A1 | 2.2a |
| Concludes a reasoned argument by stating that \((\mathbf{ABA}^{-1})^{n} = \mathbf{AB}^{n}\mathbf{A}^{-1}\) is true for \(n = 1\), and that \((\mathbf{ABA}^{-1})^{k} = \mathbf{AB}^{k}\mathbf{A}^{-1}\) implies \((\mathbf{ABA}^{-1})^{k+1} = \mathbf{AB}^{k+1}\mathbf{A}^{-1}\) and hence, by induction, that \((\mathbf{ABA}^{-1})^{n} = \mathbf{AB}^{n}\mathbf{A}^{-1}\) is true for all integers \(n \geqslant 1\) Condone \(\mathbf{A}^{-1}\mathbf{A}\) removed without reference to \(\mathbf{I}\) | R1 | 2.1 |
| (4 marks) |
Typical solution
Let \(n = 1\): \((\mathbf{ABA}^{-1})^1 = \mathbf{ABA}^{-1} = \mathbf{AB}^1\mathbf{A}^{-1}\)
\(\therefore\) it is true for \(n = 1\)
If it is true for \(n = k\), then
\[(\mathbf{ABA}^{-1})^{k} = \mathbf{AB}^{k}\mathbf{A}^{-1}\]\[\begin{aligned} &\Rightarrow (\mathbf{ABA}^{-1})^k\mathbf{ABA}^{-1} = \mathbf{AB}^k\mathbf{A}^{-1}\mathbf{ABA}^{-1} \\ &\Rightarrow (\mathbf{ABA}^{-1})^{k+1} = \mathbf{AB}^k\mathbf{IBA}^{-1} \\ &\Rightarrow (\mathbf{ABA}^{-1})^{k+1} = \mathbf{AB}^k\mathbf{BA}^{-1} \\ &\Rightarrow (\mathbf{ABA}^{-1})^{k+1} = \mathbf{AB}^{k+1}\mathbf{A}^{-1} \\ &\Rightarrow \text{it is also true for } n = k + 1\end{aligned}\]Therefore, by induction,
\[(\mathbf{ABA}^{-1})^{n} = \mathbf{AB}^{n}\mathbf{A}^{-1}\]is true for all integers \(n \geqslant 1\)