Matrices 2 (AQA A-Level Further Maths): Model Answers

Eigenvalues, Eigenvectors, and Matrix Diagonalization

Sample Answer

Part 6(a)(i): Find the value of $\lambda_1$

Working

Since $\mathbf{v}_1$ is an eigenvector with eigenvalue $\lambda_1$, we have: $M\mathbf{v}_1 = \lambda_1\mathbf{v}_1$

Substituting the given values: $\frac{1}{10}\begin{pmatrix} a & a & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} = \lambda_1\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}$

Computing the left-hand side: $\frac{1}{10}\begin{pmatrix} a + 0 - 18 \\ 0 + 0 + 0 \\ 9 + 0 - 39 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} a - 18 \\ 0 \\ -30 \end{pmatrix}$

From the third row: $\frac{-30}{10} = 3\lambda_1$ $-3 = 3\lambda_1$ $\lambda_1 = -1$

Final Answer: $\lambda_1 = -1$

Part 6 (a)(ii): Find the value of $a$

Working

From part (i), we established that $\lambda_1 = -1$ and: $\frac{1}{10}\begin{pmatrix} a - 18 \\ 0 \\ -30 \end{pmatrix} = \lambda_1\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}$

From the first row: $\frac{a - 18}{10} = \lambda_1 = -1$

Solving for $a$: $a - 18 = -10$ $a = 8$

Final Answer: $a = 8$

Part 6 (b): Find the integer $c$ and the value of $\lambda_3$

Working

Now that we know $a = 8$, the matrix is: $M = \frac{1}{10}\begin{pmatrix} 8 & 8 & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{pmatrix}$

Since $\mathbf{v}_3$ is an eigenvector with eigenvalue $\lambda_3$: $M\mathbf{v}_3 = \lambda_3\mathbf{v}_3$

$\frac{1}{10}\begin{pmatrix} 8 & 8 & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{pmatrix}\begin{pmatrix} c \\ 0 \\ 1 \end{pmatrix} = \lambda_3\begin{pmatrix} c \\ 0 \\ 1 \end{pmatrix}$

$\frac{1}{10}\begin{pmatrix} 8c - 6 \\ 0 \\ 9c - 13 \end{pmatrix} = \begin{pmatrix} c\lambda_3 \\ 0 \\ \lambda_3 \end{pmatrix}$

From the third row: $\lambda_3 = \frac{9c - 13}{10}$

From the first row: $\frac{8c - 6}{10} = c\lambda_3$

Substituting the expression for $\lambda_3$: $\frac{8c - 6}{10} = c \cdot \frac{9c - 13}{10}$

$8c - 6 = c(9c - 13)$

$8c - 6 = 9c^2 - 13c$

$0 = 9c^2 - 21c + 6$

$0 = 3c^2 - 7c + 2$

$0 = (3c - 1)(c - 2)$

This gives $c = \frac{1}{3}$ or $c = 2$

Since $c$ is an integer: $c = 2$

Therefore: $\lambda_3 = \frac{9(2) - 13}{10} = \frac{18 - 13}{10} = \frac{5}{10} = 0.5$

Final Answer: $c = 2$ and $\lambda_3 = 0.5$

Part 6(c): Find matrices $U$, $D$ and $U^{-1}$

Working

The matrix $U$ is formed using the eigenvectors as columns: $U = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 0 \\ 3 & 1 & 1 \end{pmatrix}$

The diagonal matrix $D$ contains the corresponding eigenvalues: $D = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.5 \end{pmatrix}$

To find $U^{-1}$, we calculate:

$\det(U) = 1(1 \cdot 1 - 0 \cdot 1) - 1(0 \cdot 1 - 0 \cdot 3) + 2(0 \cdot 1 - 1 \cdot 3)$ $= 1 - 0 - 6 = -5$

The adjugate matrix is: $\text{adj}(U) = \begin{pmatrix} 1 &1 & -2 \\ 0 & -5 & 0 \\ -3 & 2 & 1 \end{pmatrix}$

Therefore: $U^{-1} = \frac{1}{\det(U)}\text{adj}(U) = \frac{1}{-5}\begin{pmatrix} 1 & 1 & -2 \\ 0 & -5 & 0 \\ -3 & 2 & 1 \end{pmatrix}$

$U^{-1} = \frac{1}{5}\begin{pmatrix} -1 & -1 & 2 \\ 0 & 5 & 0 \\ 3 & -2 & -1 \end{pmatrix}$

Final Answer:

$U = \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 0 \\ 3 & 1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.5 \end{pmatrix}, \quad U^{-1} = \frac{1}{5}\begin{pmatrix} -1 & -1 & 2 \\ 0 & 5 & 0 \\ 3 & -2 & -1 \end{pmatrix}$

Marking Scheme