AS June 2019 Paper 1 Q3
3 In this question you must show detailed reasoning.
\(\mathbf{A}\) and \(\mathbf{B}\) are matrices such that \(\mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}\).
(a) Find \(\mathbf{AB}\). [3]
(b) Given that \(\mathbf{A} = \begin{pmatrix} \tfrac{1}{3} & 1 \\ 0 & 1 \end{pmatrix}\), find \(\mathbf{B}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| DR \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}\) so \(\mathbf{AB} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}^{-1}\) | M1 | 3.1a |
| \(\begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} = 3\) | B1 | 1.1 |
| \(\Rightarrow \mathbf{AB} = \tfrac{1}{3}\begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}\) | A1cao | 1.1 |
| [3] |
Notes
M1: \(\mathbf{AB} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}^{-1}\)
B1: soi
A1cao: or \(\begin{pmatrix} \tfrac{1}{3} & -\tfrac{1}{3} \\ \tfrac{1}{3} & \tfrac{2}{3} \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathbf{A}^{-1} = 3\begin{pmatrix} 1 & -1 \\ 0 & \tfrac{1}{3} \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{B} = \mathbf{A}^{-1}\mathbf{AB} = 3 \times \tfrac{1}{3}\begin{pmatrix} 1 & -1 \\ 0 & \tfrac{1}{3} \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}\) | M1 | 3.1a |
| \(= \begin{pmatrix} 0 & -3 \\ \tfrac{1}{3} & \tfrac{2}{3} \end{pmatrix}\) | A1cao | 1.1 |
| [3] |
Notes
M1: pre-multiply their \(\mathbf{AB}\) by \(\mathbf{A}^{-1}\)
A1cao: Or \(\tfrac{1}{3}\begin{pmatrix} 0 & -9 \\ 1 & 2 \end{pmatrix}\)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\mathbf{B}^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{A} = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} \tfrac{1}{3} & 1 \\ 0 & 1 \end{pmatrix}\) | M1 |
| \(= \begin{pmatrix} \tfrac{2}{3} & 3 \\ -\tfrac{1}{3} & 0 \end{pmatrix}\) | A1 |
| \(\Rightarrow \mathbf{B} = \begin{pmatrix} 0 & -3 \\ \tfrac{1}{3} & \tfrac{2}{3} \end{pmatrix}\) | A1cao |
M1: post-multiply \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) by \(\mathbf{A}\)
A1cao: Or \(\tfrac{1}{3}\begin{pmatrix} 0 & -9 \\ 1 & 2 \end{pmatrix}\)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} \tfrac{1}{3} & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \tfrac{1}{3} & -\tfrac{1}{3} \\ \tfrac{1}{3} & \tfrac{2}{3} \end{pmatrix}\) \(\Rightarrow \dfrac{1}{3}a + c = \dfrac{1}{3},\ \dfrac{1}{3}b + d = -\dfrac{1}{3}\) | M1 |
| \(c = \dfrac{1}{3},\ d = \dfrac{2}{3}\) | A1 |
| \(\Rightarrow a = 0,\ b = -3 \Rightarrow \mathbf{B} = \begin{pmatrix} 0 & -3 \\ \tfrac{1}{3} & \tfrac{2}{3} \end{pmatrix}\) | A1 |
M1: forming equations (any two)
ft their \(\mathbf{AB}\)