FP3 June 2013 (R) Q6
6. It is given that \(\begin{pmatrix}1 \\ 2 \\ 0\end{pmatrix}\) is an eigenvector of the matrix \(\mathbf{A}\), where \[\mathbf{A} = \begin{pmatrix}4 & 2 & 3 \\ 2 & b & 0 \\ a & 1 & 8\end{pmatrix}\] and \(a\) and \(b\) are constants.
(a) Find the eigenvalue of \(\mathbf{A}\) corresponding to the eigenvector \(\begin{pmatrix}1 \\ 2 \\ 0\end{pmatrix}\). (3)
(b) Find the values of \(a\) and \(b\). (3)
(c) Find the other eigenvalues of \(\mathbf{A}\). (5)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}4 & 2 & 3 \\ 2 & b & 0 \\ a & 1 & 8\end{pmatrix}\begin{pmatrix}1 \\ 2 \\ 0\end{pmatrix} = \begin{pmatrix}8 \\ \ldots \\ \ldots\end{pmatrix},\ = \lambda\begin{pmatrix}1 \\ 2 \\ 0\end{pmatrix},\ \lambda = 8\) M1: Multiplies the given matrix by the given eigenvector M1: Equates the \(x\) value to \(\lambda\) A1: \(\lambda = 8\) | M1, M1, A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}8 \\ 2 + 2b \\ a + 2\end{pmatrix} = \text{"}8\text{"}\begin{pmatrix}1 \\ 2 \\ 0\end{pmatrix}\) So \(a = -2\) and \(b = 7\) M1: Their \(2 + 2b = 2\lambda\) or their \(a + 2 = 0\) A1: \(b = 7\) or \(a = -2\) A1: \(b = 7\) and \(a = -2\) | M1 A1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}4 - \lambda & 2 & 3 \\ 2 & 7 - \lambda & 0 \\ -2 & 1 & 8 - \lambda\end{vmatrix} = 0\) \(\therefore (4 - \lambda)(7 - \lambda)(8 - \lambda) - 2 \times 2(8 - \lambda) + 3(2 + 2(7 - \lambda)) = 0\) Correct attempt to establish the Characteristic Equation. = 0 is required but may be implied by later work Allow this mark if the equation is in terms of a, b, c | M1 |
| Attempts to factorise i.e. \((8 - \lambda)(30 - 11\lambda + \lambda^2)\) or \((6 - \lambda)(40 - 13\lambda + \lambda^2)\) or \((5 - \lambda)(48 - 14\lambda + \lambda^2)\) (NB \(240 - 118\lambda + 19\lambda^2 - \lambda^3 = 0\)) M1: Attempt to factorise their cubic – an attempt to identify a linear factor and processes to obtain a simplified quadratic factor A1: Correct factorisation into one linear and one quadratic factor | M1 A1 |
| Eigenvalues are 5 and 6 M1: Solves their equation to obtain the other eigenvalues A1: 5 and 6 | M1 A1 |
| (5) | |
| (11 marks) |
Notes
(corrected from the printed mark scheme: the question total is printed as “Total 8”; the parts are worth 3 + 3 + 5 = 11 marks, as on the question paper)