A2 June 2023 Paper 2 Q8
8 \(\mathbf{A}\) is a non-singular \(2 \times 2\) matrix and \(\mathbf{A}^{\mathrm{T}}\) is the transpose of \(\mathbf{A}\)
(a) Using the result\[(\mathbf{AB})^{\mathrm{T}} = \mathbf{B}^{\mathrm{T}}\mathbf{A}^{\mathrm{T}}\]
show that
\[\left(\mathbf{A}^{-1}\right)^{\mathrm{T}} = \left(\mathbf{A}^{\mathrm{T}}\right)^{-1}\] [3 marks](b) It is given that \(\mathbf{A} = \begin{bmatrix} 4 & 5 \\ -1 & k \end{bmatrix}\), where \(k\) is a real constant.
(i) Find \(\left(\mathbf{A}^{-1}\right)^{\mathrm{T}}\), giving your answer in terms of \(k\) [2 marks]
(ii) State the restriction on the possible values of \(k\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Uses the result \((\mathbf{AB})^{\mathrm{T}} = \mathbf{B}^{\mathrm{T}}\mathbf{A}^{\mathrm{T}}\) in a statement involving \(\mathbf{A}^{-1}\) Use of the notation \(\mathbf{A}^{-\mathrm{T}}\) is not acceptable here. | M1 | 3.1a |
| Uses the fact that the identity matrix is its own transpose. PI | M1 | 1.1a |
| Completes a reasoned argument to show the required result. | R1 | 2.1 |
| (3) |
Typical solution
\[\left(\mathbf{A}^{\mathrm{T}}\right)\left(\mathbf{A}^{-1}\right)^{\mathrm{T}} = \left(\mathbf{A}^{-1}\mathbf{A}\right)^{\mathrm{T}} = \mathbf{I}^{\mathrm{T}} = \mathbf{I}\]\[\therefore \left(\mathbf{A}^{\mathrm{T}}\right)^{-1} = \left(\mathbf{A}^{-1}\right)^{\mathrm{T}}\]| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(\mathbf{A}^{-1}\) or \(\mathbf{A}^{\mathrm{T}}\) | M1 | 1.1a |
| (i) Obtains \(\left(\mathbf{A}^{-1}\right)^{\mathrm{T}}\) Allow answer with factor outside matrix. | A1 | 1.1b |
| (2) | ||
| (ii) Obtains correct restriction on \(k\) FT their \(\det(\mathbf{A})\) from (b)(i). | B1F | 1.1b |
| (1) | ||
| (6 marks) |
Typical solution
(i)
\[\mathbf{A}^{-1} = \frac{1}{4k + 5}\begin{bmatrix} k & -5 \\ 1 & 4 \end{bmatrix}\]\[\left(\mathbf{A}^{-1}\right)^{\mathrm{T}} = \begin{bmatrix} \dfrac{k}{4k + 5} & \dfrac{1}{4k + 5} \\[3ex] \dfrac{-5}{4k + 5} & \dfrac{4}{4k + 5} \end{bmatrix}\](ii)
\[k \neq -\frac{5}{4}\]