FP1 January 2012 Q8
8. \[\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}\]
(a) Show that \(\mathbf{A}\) is non-singular. (2)
(b) Find \(\mathbf{B}\) such that \(\mathbf{BA}^2 = \mathbf{A}\). (4)
| Scheme | Marks |
|---|---|
| \(\det(\mathbf{A}) = 3 \times 0 - 2 \times 1\ (= -2)\) Correct attempt at the determinant | M1 |
| \(\det(\mathbf{A}) \neq 0\) (so \(\mathbf{A}\) is non singular) \(\det(\mathbf{A}) = -2\) and some reference to zero | A1 |
| (2) |
Notes
\(\dfrac{1}{\det(\mathbf{A})}\) scores M0
| Scheme | Marks |
|---|---|
| \(\mathbf{BA}^2 = \mathbf{A} \Rightarrow \mathbf{BA} = \mathbf{I} \Rightarrow \mathbf{B} = \mathbf{A}^{-1}\) Recognising that \(\mathbf{A}^{-1}\) is required | M1 |
| \(\mathbf{B} = -\dfrac{1}{2}\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}\) At least 3 correct terms in \(\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}\) \(\dfrac{1}{\text{their }\det(\mathbf{A})}\begin{pmatrix} * & * \\ * & * \end{pmatrix}\) Fully correct answer | M1 B1ft A1 |
| (4) | |
| (6 marks) |
Notes
Correct answer only score 4/4
Ignore poor matrix algebra notation if the intention is clear
Alternative ((b) Way 2)
| Scheme | Marks |
|---|---|
| \(\mathbf{A}^2 = \begin{pmatrix} 2 & 3 \\ 6 & 11 \end{pmatrix}\) Correct matrix | B1 |
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 6 & 11 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix} \Rightarrow \begin{matrix} 2a + 6b = 0 \\ 3a + 11b = 1 \end{matrix}\ \text{or}\ \begin{matrix} 2c + 6d = 2 \\ 3c + 11d = 3 \end{matrix}\) 2 equations in \(a\) and \(b\) or 2 equations in \(c\) and \(d\) | M1 |
| \(a = -\dfrac{3}{2}, b = \dfrac{1}{2}, c = 1, d = 0\) M1 Solves for \(a\) and \(b\) or \(c\) and \(d\) A1 All 4 values correct | M1A1 |
Alternative ((b) Way 3)
| Scheme | Marks |
|---|---|
| \(\mathbf{A}^2 = \begin{pmatrix} 2 & 3 \\ 6 & 11 \end{pmatrix}\) Correct matrix | B1 |
| \(\left(\mathbf{A}^2\right)^{-1} = \dfrac{1}{\text{"}2\text{"} \times \text{"}11\text{"} - \text{"}3\text{"} \times \text{"}6\text{"}}\begin{pmatrix} \text{"}11\text{"} & \text{"}{-3}\text{"} \\ \text{"}{-6}\text{"} & \text{"}2\text{"} \end{pmatrix}\) see note Attempt inverse of \(\mathbf{A}^2\) | M1 |
| \(\mathbf{A}\left(\mathbf{A}^2\right)^{-1} = \dfrac{1}{4}\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 11 & -3 \\ -6 & 2 \end{pmatrix}\ \text{or}\ \dfrac{1}{4}\begin{pmatrix} 11 & -3 \\ -6 & 2 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}\) Attempts \(\mathbf{A}\left(\mathbf{A}^2\right)^{-1}\) or \(\left(\mathbf{A}^2\right)^{-1}\mathbf{A}\) | M1 |
| \(\mathbf{B} = -\dfrac{1}{2}\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}\) Fully correct answer | A1 |
8(b) Way 3 note: Attempting inverse of \(\mathbf{A}^2\) needs to be recognisable as an attempt at an inverse
E.g. \(\left(\mathbf{A}^2\right)^{-1} = \dfrac{1}{\textit{Their Det}(\mathbf{A}^2)}\left(\text{A } \textit{changed } \mathbf{A}^2\right)\)
Alternative ((b) Way 4)
| Scheme | Marks |
|---|---|
| \(\mathbf{BA} = \mathbf{I}\) Recognising that \(\mathbf{BA} = \mathbf{I}\) | B1 |
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Rightarrow \begin{matrix} 2b = 1 \\ a + 3b = 0 \end{matrix}\ \text{or}\ \begin{matrix} 2d = 0 \\ c + 3d = 1 \end{matrix}\) 2 equations in \(a\) and \(b\) or 2 equations in \(c\) and \(d\) | M1 |
| \(a = -\dfrac{3}{2}, b = \dfrac{1}{2}, c = 1, d = 0\) M1 Solves for \(a\) and \(b\) or \(c\) and \(d\) A1 All 4 values correct | M1A1 |