FP3 June 2013 Q5
5. The matrix \(\mathbf{M}\) is given by \[\mathbf{M} = \begin{pmatrix}1 & 1 & a \\ 2 & b & c \\ -1 & 0 & 1\end{pmatrix}, \quad \text{where } a, b \text{ and } c \text{ are constants.}\]
(a) Given that \(\mathbf{j} + \mathbf{k}\) and \(\mathbf{i} - \mathbf{k}\) are two of the eigenvectors of \(\mathbf{M}\),
find
find
(i) the values of \(a\), \(b\) and \(c\),
(ii) the eigenvalues which correspond to the two given eigenvectors.
(8)(b) The matrix \(\mathbf{P}\) is given by \[\mathbf{P} = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & d \\ -1 & 0 & 1\end{pmatrix}, \quad \text{where } d \text{ is constant, } d \neq -1\] Find
(i) the determinant of \(\mathbf{P}\) in terms of \(d\),
(ii) the matrix \(\mathbf{P}^{-1}\) in terms of \(d\).
(5)| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}1 & 1 & a \\ 2 & b & c \\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} = \begin{pmatrix}1 + a \\ b + c \\ 1\end{pmatrix} = \lambda_1\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\), and so \(a = -1,\ \lambda_1 = 1\) M1: Multiplies out matrix with first eigenvector and puts equal to \(\lambda_1\) times eigenvector. A1: Deduces \(a = -1\). A1: Deduces \(\lambda_1 = 1\) | M1, A1, A1 |
| \(\begin{pmatrix}1 & 1 & a \\ 2 & b & c \\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix}1 - a \\ 2 - c \\ -2\end{pmatrix} = \lambda_2\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\), and so \(c = 2,\ \lambda_2 = 2\) M1: Multiplies out matrix with second eigenvector and puts equal to \(\lambda_2\) times eigenvector. A1: Deduces \(c = 2\). A1: Deduces \(\lambda_2 = 2\) | M1, A1, A1 |
| \(b + c = \lambda_1\) so \(b = -1\) M1: Uses \(b + c = \lambda_1\) with their \(\lambda_1\) to find a value for \(b\) (They must have an equation in \(b\) and \(c\) from the first eigenvector to score this mark) A1: \(b = -1\) | M1A1 |
| \((a = -1,\ b = -1,\ c = 2,\ \lambda_1 = 1,\ \lambda_2 = 2)\) | |
| (8) |
| Scheme | Marks |
|---|---|
| (i) \(\det\mathbf{P} = -d - 1\) Allow \(1 - d - 2\) or \(1 - (2 + d)\) A correct (possibly un-simplified) determinant | B1 |
| (ii) \(\mathbf{P}^T = \begin{pmatrix}1 & 2 & -1 \\ 1 & 1 & 0 \\ 0 & d & 1\end{pmatrix}\) or minors \(\begin{pmatrix}1 & d+2 & 1 \\ 1 & 1 & 1 \\ d & d & -1\end{pmatrix}\) or cofactors \(\begin{pmatrix}1 & -2-d & 1 \\ -1 & 1 & -1 \\ d & -d & -1\end{pmatrix}\) a correct first step | B1 |
| \(\dfrac{1}{-d - 1}\begin{pmatrix}1 & -1 & d \\ -2-d & 1 & -d \\ 1 & -1 & -1\end{pmatrix}\) M1: Identifiable full attempt at inverse including reciprocal of determinant. Could be indicated by at least 6 correct elements. A1: Two rows or two columns correct (ignoring determinant) BUT M0A1A0 or M0A1A1 is not possible A1: Fully correct inverse | M1 A1 A1 |
| (5) | |
| (13 marks) |