A2 June 2023 Paper 1 Q6
6 The matrix \(\mathbf{M}\) is given by
\[\mathbf{M} = \frac{1}{10}\begin{bmatrix} a & a & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{bmatrix}\]where \(a\) is a real number.
The vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are eigenvectors of \(\mathbf{M}\)
The corresponding eigenvalues are \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\) respectively.
It is given that \(\lambda_2 = 1\) and \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}\), \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_3 = \begin{bmatrix} c \\ 0 \\ 1 \end{bmatrix}\),
where \(c\) is an integer.
(a)
(i) Find the value of \(\lambda_1\) [2 marks]
(ii) Find the value of \(a\) [2 marks]
(b) Find the integer \(c\) and the value of \(\lambda_3\) [4 marks]
(c) Find matrices \(\mathbf{U}\), \(\mathbf{D}\) and \(\mathbf{U}^{-1}\), such that \(\mathbf{D}\) is diagonal and \(\mathbf{M} = \mathbf{UDU}^{-1}\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Uses an appropriate method to obtain the value of \(\lambda_1\) | M1 | 1.1a |
| (i) Obtains the correct value of \(\lambda_1\) | A1 | 1.1b |
| (2) | ||
| (ii) Uses eigenvector definition to set up a linear equation in \(a\) | M1 | 1.1a |
| (ii) Deduces the correct value of \(a\) | A1 | 2.2a |
| (2) |
Typical solution
(i)
\[\frac{1}{10}\begin{bmatrix} a & a & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} a - 18 \\ 0 \\ -30 \end{bmatrix} = \lambda_1\begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}\]\[\lambda_1 = -1\](ii)
\[10 = 18 - a\]\[a = 8\]| Scheme | Marks | AO |
|---|---|---|
| Uses an appropriate method to find \(\lambda\) or \(c\) PI by \(\lambda_3 = 0.5\) | M1 | 3.1a |
| Forms an equation in \(\lambda\) or \(c\) PI by \(\lambda_3 = 0.5\) | M1 | 1.1a |
| Finds the correct \(\lambda_3\) | A1 | 1.1b |
| Deduces the correct \(c\) NMS = 0/4 if \(\lambda_3\) wrong | A1 | 2.2a |
| (4) |
Typical solution
\[\frac{1}{10}\begin{bmatrix} 8 & 8 & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{bmatrix}\begin{bmatrix} c \\ 0 \\ 1 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 8c - 6 \\ 0 \\ 9c - 13 \end{bmatrix} = \lambda_3\begin{bmatrix} c \\ 0 \\ 1 \end{bmatrix}\]\[\tfrac{1}{10}(8c - 6) = \lambda_3 c \quad\text{and}\quad \tfrac{1}{10}(9c - 13) = \lambda_3\]\[\tfrac{1}{10}(8c - 6) = \tfrac{c}{10}(9c - 13)\]\[9c^2 - 21c + 6 = 0\]\[c = 2 \text{ and } c = \tfrac{1}{3}\ (\text{reject as } c \in \mathbb{Z})\]\[c = 2 \text{ and } \lambda_3 = 0.5\]| Scheme | Marks | AO |
|---|---|---|
| Deduces a correct \(\mathbf{U}\), ft their \(c\) only. Condone “\(c\)” | B1F | 2.2a |
| Deduces the value of \(\mathbf{D}\), ft their \(\lambda_1\) and \(\lambda_3\). Must be compatible with their \(\mathbf{U}\). Condone “\(\lambda_1\)” and “\(\lambda_3\)” | B1F | 2.2a |
| Obtains \(\mathbf{U}^{-1}\), ft their \(\mathbf{U}\) | B1F | 1.1b |
| (3) | ||
| (11 marks) |