A2 June 2023 Paper 1 Q6

AQACurrent spec11 marksMatrices

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]

Mark scheme (a)

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)

\[10 = 18 - a\]\[a = 8\]

Mark scheme (b)

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

\[\frac{1}{10}\begin{bmatrix} 8 & 8 & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{bmatrix}\begin{bmatrix} c \\ 0 \\ 1 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 8c - 6 \\ 0 \\ 9c - 13 \end{bmatrix} = \lambda_3\begin{bmatrix} c \\ 0 \\ 1 \end{bmatrix}\]\[\tfrac{1}{10}(8c - 6) = \lambda_3 c \quad\text{and}\quad \tfrac{1}{10}(9c - 13) = \lambda_3\]\[\tfrac{1}{10}(8c - 6) = \tfrac{c}{10}(9c - 13)\]\[9c^2 - 21c + 6 = 0\]\[c = 2 \text{ and } c = \tfrac{1}{3}\ (\text{reject as } c \in \mathbb{Z})\]\[c = 2 \text{ and } \lambda_3 = 0.5\]

Mark scheme (c)

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)

Typical solution

\[\mathbf{U} = \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 0 \\ 3 & 1 & 1 \end{bmatrix}\]\[\mathbf{D} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.5 \end{bmatrix}\]\[\mathbf{U}^{-1} = \frac{1}{5}\begin{bmatrix} -1 & -1 & 2 \\ 0 & 5 & 0 \\ 3 & -2 & -1 \end{bmatrix}\]