A2 June 2019 Paper 1 Q3
3 Matrices \(\mathbf{A}\) and \(\mathbf{B}\) are defined by \(\mathbf{A} = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} k & 1 \\ 2 & 0 \end{pmatrix}\), where \(k\) is a constant.
(a) Verify the result \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) in this case. [5]
(b) Investigate whether \(\mathbf{A}\) and \(\mathbf{B}\) are commutative under matrix multiplication. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} k & 1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 3k + 2 & 3 \\ 2k + 2 & 2 \end{pmatrix}\) | B1 | 1.1b |
| \((\mathbf{AB})^{-1} = -\dfrac{1}{2}\begin{pmatrix} 2 & -3 \\ -2k - 2 & 3k + 2 \end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{A}^{-1} = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix}\) | B1 | 1.1b |
| \(\mathbf{B}^{-1} = -\dfrac{1}{2}\begin{pmatrix} 0 & -1 \\ -2 & k \end{pmatrix}\) | B1 | 1.1b |
| \(\mathbf{B}^{-1}\mathbf{A}^{-1} = -\dfrac{1}{2}\begin{pmatrix} 2 & -3 \\ -2k - 2 & 3k + 2 \end{pmatrix}\) [so \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)] | B1 | 2.2a |
| [5] |
Notes
B1ft: ft their \(\mathbf{AB}\) provided det \(\neq 0\) [isw]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{BA} = \begin{pmatrix} 3k + 2 & k + 1 \\ 6 & 2 \end{pmatrix}\) | B1 | 1.1b |
| \(\mathbf{AB} = \mathbf{BA}\) when \(k = 2\) [and not otherwise] | B1 | 2.3 |
| [2] |