AS October 2021 Paper 1 Q4
4 Anika thinks that, for two square matrices \(\mathbf{A}\) and \(\mathbf{B}\), the inverse of \(\mathbf{AB}\) is \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). Her attempted proof of this is as follows.
\[\begin{aligned} (\mathbf{AB})(\mathbf{A}^{-1}\mathbf{B}^{-1}) &= \mathbf{A}(\mathbf{BA}^{-1})\mathbf{B}^{-1} \\ &= \mathbf{A}(\mathbf{A}^{-1}\mathbf{B})\mathbf{B}^{-1} \\ &= (\mathbf{AA}^{-1})(\mathbf{BB}^{-1}) \\ &= \mathbf{I} \times \mathbf{I} \\ &= \mathbf{I} \end{aligned}\]Hence \((\mathbf{AB})^{-1} = \mathbf{A}^{-1}\mathbf{B}^{-1}\)
(a) Explain the error in Anika’s working. [2]
(b) State the correct inverse of the matrix \(\mathbf{AB}\) and amend Anika’s working to prove this. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{BA}^{-1} \ne \mathbf{A}^{-1}\mathbf{B}\) | M1 | 2.3 |
| as matrix multiplication is not commutative | A1 | 2.4 |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| Correct inverse of \(\mathbf{AB}\) is \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) | B1 | 2.1 |
| \((\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) = \mathbf{A}(\mathbf{BB}^{-1})\mathbf{A}^{-1}\) | M1 | 2.1 |
| \([= \mathbf{AIA}^{-1}] = \mathbf{AA}^{-1} = \mathbf{I}\) | A1 | 2.2a |
| [3] |
Notes
B1: SCB1 for a correct proof but not using Anika’s working
M1: \(\mathbf{BB}^{-1} = \mathbf{I}\) soi [need not show \(\mathbf{I}\)]
A1: \(\ldots\mathbf{AA}^{-1} = \mathbf{I}\)