AS June 2025 Paper 1 Q1
1 The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by
\(\mathbf{A} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}\).
(a) Use \(\mathbf{A}\) and \(\mathbf{B}\) to show that matrix multiplication is not, in general, commutative. [2]
(b) Verify that \(\mathbf{A}\) and \(\mathbf{B}\) satisfy \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 0 & 1 \end{pmatrix}\) \(\mathbf{BA} = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 7 \\ 0 & 1 \end{pmatrix}\) | B1 | 1.1 |
| so \(\mathbf{AB} \ne \mathbf{BA}\) | B1 | 2.1 |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{AB})^{-1} = \dfrac{1}{2}\begin{pmatrix} 1 & -4 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -2 \\ 0 & 1 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{A}^{-1} = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}\quad \mathbf{B}^{-1} = \dfrac{1}{2}\begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{B}^{-1}\mathbf{A}^{-1} = \dfrac{1}{2}\begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} 1 & -4 \\ 0 & 2 \end{pmatrix} = (\mathbf{AB})^{-1}\) | B1 | 2.1 |
| [3] |
Notes
B1: both correct
B1: must have a final statement of equality