A2 June 2023 Q1
1.
\[\mathbf{A} = \begin{pmatrix} -1 & a \\ 3 & 8 \end{pmatrix}\]where \(a\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| \(\det\begin{pmatrix} -1-\lambda & a \\ 3 & 8-\lambda \end{pmatrix} = 0 \Rightarrow (-1-\lambda)(8-\lambda) - 3a = 0 \Rightarrow \ldots\) | M1 | 1.1b |
| \(\lambda^2 - 7\lambda - 3a - 8 = 0\) o.e. | A1 | 1.1b |
| (2) |
Notes
M1: Correct method to find the characteristic equation for \(\mathbf{A}\), condone missing = 0
A1: Correct simplified characteristic equation.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A}^2 - 7\mathbf{A} - 3a\mathbf{I} - 8\mathbf{I} = 0 \Rightarrow \mathbf{A}^3 = 7\mathbf{A}^2 + (3a + 8)\mathbf{A}\) | M1 | 1.1b |
| \(\Rightarrow \mathbf{A}^3 = 7\big(7\mathbf{A} + (3a + 8)\mathbf{I}\big) + (3a + 8)\mathbf{A}\) Or \(\mathbf{A}^3 = 7\begin{pmatrix} -1 & a \\ 3 & 8 \end{pmatrix}^2 + (8 + 3a)\begin{pmatrix} -1 & a \\ 3 & 8 \end{pmatrix}\) \(= 7\begin{pmatrix} 1+3a & 7a \\ 21 & 3a+64 \end{pmatrix} + \begin{pmatrix} -8-3a & 8a+3a^2 \\ 24+9a & 64+24a \end{pmatrix}\) | M1 | 2.1 |
| \(\Rightarrow \mathbf{A}^3 = (3a + 8 + 49)\mathbf{A} + 7(3a + 8)\mathbf{I} \Rightarrow 3a + 57 = 1 \Rightarrow a = \ldots\) Or \(\begin{pmatrix} b-1 & a \\ 3 & b+8 \end{pmatrix} = \begin{pmatrix} 18a-1 & 3a^2+57a \\ 171+9a & 512+45a \end{pmatrix} \Rightarrow\) e.g. \(3 = 171 + 9a \Rightarrow a = \ldots\) Or \(\mathbf{A}^3 = \begin{pmatrix} -1 & a \\ 3 & 8 \end{pmatrix} + \begin{pmatrix} 18a & 3a^2+56a \\ 168+9a & 504+45a \end{pmatrix} \Rightarrow\) e.g. \(18a = 504 + 45a \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(\Rightarrow a = -\dfrac{56}{3},\ b = -336\) | A1 | 1.1b |
| (4) | ||
| (6 marks) |
Notes
Note: this question asks the candidates to use the Cayley-Hamilton theorem so any other approach that doesn’t score the first method mark scores no marks
M1: Uses the Cayley-Hamilton theorem with their equation and multiplies though by \(\mathbf{A}\) to find an equation for \(\mathbf{A}^3\)
M1: Substitutes for \(\mathbf{A}^2\) to obtain an equation for \(\mathbf{A}^3\) in terms of \(a\), \(\mathbf{I}\) and \(\mathbf{A}\). Alternatively substitutes in the matrix \(\mathbf{A}\) and attempts to square
M1: Equates coefficient(s) of \(\mathbf{A}\) to 1 and proceeds to find a value for \(a\). Alternatively equates elements to find a value for \(a\).
A1: Correct values for \(a\) and \(b\).
Special case: Missing matrix \(\mathbf{I}\) from their working can score maximum of M1 (if multiply by A correctly) M1M1A0 unless implied \(\mathbf{I}\) from their working this can score all marks.
(corrected from the printed mark scheme: in the two “Or” methods for the third mark the bottom-right entries are printed as \(512 + 27a\) and \(504 + 27a\), and the example as \(18a = 504 + 27a\); the correct entries are \(512 + 45a\) and \(504 + 45a\))