A2 June 2025 Paper 1 Q4
4
Write down the product rule for the inverse matrices of \(\mathbf{M}\), \(\mathbf{N}\) and \(\mathbf{MN}\). [1]
\(\mathbf{M} = \begin{pmatrix} a & 1 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{N} = \begin{pmatrix} 0 & -1 \\ 1 & b \end{pmatrix}\) and \(a\) and \(b\) are non-zero constants. [6]
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{MN})^{-1} = \mathbf{N}^{-1}\mathbf{M}^{-1}\) | B1 | 1.2 |
| [1] |
Notes
B1: do not allow misread of \(\mathbf{NM}\) for \(\mathbf{MN}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{M} = a\) | B1 | 1.1 |
| \(\mathbf{M}^{-1} = \frac{1}{a}\begin{pmatrix} 1 & -1 \\ 0 & a \end{pmatrix}\) | B1 | 2.1 |
| \(\mathbf{N}^{-1} = \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{MN} = \begin{pmatrix} 1 & b - a \\ 1 & b \end{pmatrix}\) | B1 | 1.1 |
| \((\mathbf{MN})^{-1} = \frac{1}{a}\begin{pmatrix} b & a - b \\ -1 & 1 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{N}^{-1}\mathbf{M}^{-1} = \frac{1}{a}\begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & a \end{pmatrix} = \frac{1}{a}\begin{pmatrix} b & a - b \\ -1 & 1 \end{pmatrix}\) | B1 | 2.2a |
| [6] |
Notes
B1: it must be unambiguous which matrix is which for each mark to be awarded
B1 (\(\mathbf{MN}\)): Allow \(\mathbf{NM} = \begin{pmatrix} 0 & -1 \\ a & b + 1 \end{pmatrix}\) if \((\mathbf{NM})^{-1} = \mathbf{M}^{-1}\mathbf{N}^{-1}\) given in part (a)
B1 (\((\mathbf{MN})^{-1}\)): may be unsimplified. Allow \((\mathbf{NM})^{-1} = \frac{1}{a}\begin{pmatrix} b + 1 & 1 \\ -a & 0 \end{pmatrix}\) if \((\mathbf{NM})^{-1} = \mathbf{M}^{-1}\mathbf{N}^{-1}\) given in part (a)
B1 (\(\mathbf{N}^{-1}\mathbf{M}^{-1}\)): product of two matrices must be seen. Allow \(\mathbf{M}^{-1}\mathbf{N}^{-1} = \frac{1}{a}\begin{pmatrix} 1 & -1 \\ 0 & a \end{pmatrix}\begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix} = \frac{1}{a}\begin{pmatrix} b + 1 & 1 \\ -a & 0 \end{pmatrix}\) if \((\mathbf{NM})^{-1} = \mathbf{M}^{-1}\mathbf{N}^{-1}\) given in part (a).