A2 June 2025 Paper 1 Q12
12
(a) Find the eigenvalues and corresponding eigenvectors of the matrix\[\mathbf{M} = \frac{1}{20}\begin{bmatrix} 19 & 3 \\ 3 & 11 \end{bmatrix}\] [5 marks]
(b) State, with a reason, the Cartesian equation of the line of invariant points of the matrix \(\mathbf{M}\) [2 marks]
(c) Find matrices \(\mathbf{U}\), \(\mathbf{D}\) and \(\mathbf{U}^{-1}\), such that \(\mathbf{D}\) is diagonal and \(\mathbf{M} = \mathbf{U}\mathbf{D}\mathbf{U}^{-1}\) [3 marks]
(d) Hence, find the matrix \(\mathbf{L}\) such that \(\mathbf{M}^n \to \mathbf{L}\) as \(n \to \infty\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms correct characteristic equation and solves. PI by \(\lambda = 1\) & \(\lambda = 0.5\) Condone \(\lambda = 20\) & \(\lambda = 10\) from \(0 = (19 - \lambda)(11 - \lambda) - 9\) | M1 | 1.1a |
| Obtains \(\lambda = 1\) & \(\lambda = 0.5\) | A1 | 1.1b |
| Uses correct equation to find eigenvector for one of their two eigenvalues. PI by a correct eigenvector. | M1 | 1.1a |
| Obtains a correct eigenvector for one of their two eigenvalues. Allow any scalar multiple. | A1 | 1.1b |
| Obtains both correct eigenvectors paired with the corresponding correct eigenvalue. Allow any scalar multiple. | A1 | 1.1b |
| (5) |
Typical solution
\[0 = \left(\frac{19}{20} - \lambda\right)\left(\frac{11}{20} - \lambda\right) - \frac{9}{400}\]\[0 = 2\lambda^2 - 3\lambda + 1\]\[\lambda = 1 \;\&\; \lambda = 0.5\]\[\lambda = 1 : \mathbf{0} = \begin{bmatrix} \frac{-1}{20} & \frac{3}{20} \\[4pt] \frac{3}{20} & \frac{-9}{20} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\]\[\lambda = 1 : \begin{bmatrix} 3 \\ 1 \end{bmatrix}\]\[\lambda = 0.5 : \mathbf{0} = \begin{bmatrix} \frac{9}{20} & \frac{3}{20} \\[4pt] \frac{3}{20} & \frac{1}{20} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\]\[\lambda = 0.5 : \begin{bmatrix} -1 \\ 3 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Explains that for a line of invariant points \(\lambda = 1\) or \(\mathbf{Mv} = \mathbf{v}\) | M1 | 2.4 |
| Deduces \(y = \dfrac{1}{3}x\) OE from \(\lambda = 1\) | R1 | 2.2a |
| (2) |
Typical solution
As \(\lambda = 1\) for a line of invariant points.
\[y = \frac{1}{3}x\]| Scheme | Marks | AO |
|---|---|---|
| Deduces their correct \(\mathbf{U}\) with no zero column. | B1F | 2.2a |
| Deduces their correct \(\mathbf{D}\) Must be compatible with their \(\mathbf{U}\) | B1F | 2.2a |
| Obtains \(\mathbf{U}^{-1}\), ft their \(\mathbf{U}\) | B1F | 1.1b |
| (3) |
Typical solution
\[\mathbf{U} = \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix}\]\[\mathbf{D} = \begin{bmatrix} 1 & 0 \\ 0 & 0.5 \end{bmatrix}\]\[\mathbf{U}^{-1} = \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Multiplies their matrices in correct order with powers inside \(\mathbf{D}^n\) | M1 | 1.1a |
| Deduces that as \(n \to \infty\) limit of \(\mathbf{D}^n\) is \(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\) or \(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\) | B1 | 2.2a |
| Completes a reasoned argument to obtain correct \(\mathbf{L}\) | R1 | 2.1 |
| (3) | ||
| (13 marks) |