AS June 2019 Q1

EdexcelCurrent spec5 marksFurther Matrices

1. Given that

\[\mathbf{A} = \begin{pmatrix} 3 & 2 \\ 2 & 2 \end{pmatrix}\]

(a) find the characteristic equation for the matrix \(\mathbf{A}\), simplifying your answer. (2)

(b) Hence find an expression for the matrix \(\mathbf{A}^{-1}\) in the form \(\lambda\mathbf{A} + \mu\mathbf{I}\), where \(\lambda\) and \(\mu\) are constants to be found. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\det\begin{pmatrix} 3-\lambda & 2 \\ 2 & 2-\lambda \end{pmatrix} = (3-\lambda)(2-\lambda) - 4\,(= 0)\)	M1	1.1b
\(\lambda^2 - 5\lambda + 2 = 0\)	A1	1.1b
	(2)

Notes

M1: Complete method to find the characteristic equation, condone missing = 0

A1: Obtains a correct three term quadratic equation – may use any variable.

Mark scheme (b)

Scheme	Marks	AO
\(\mathbf{A}^2 - 5\mathbf{A} + 2\mathbf{I} = 0\)	B1ft	1.1b
Multiplies through by \(\mathbf{A}^{-1}\) \(\mathbf{A} - 5\mathbf{I} + 2\mathbf{A}^{-1} = 0\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\) OR Rearranges to make \(\mathbf{I}\) the subject, takes out a factor of \(\mathbf{A}\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\) \(\mathbf{I} = \dfrac{(5\mathbf{A} - \mathbf{A}^2)}{2} = \mathbf{A}\dfrac{(5\mathbf{I} - \mathbf{A})}{2} \Rightarrow \mathbf{A}^{-1} = \ldots\) OR Rearranges to make \(\mathbf{I}\) the subject and multiplies through by \(\mathbf{A}^{-1}\) \(\mathbf{I} = \dfrac{5}{2}\mathbf{A} - \dfrac{1}{2}\mathbf{A}^2 \Rightarrow \mathbf{A}^{-1} = \dfrac{5}{2}\mathbf{A}\mathbf{A}^{-1} - \dfrac{1}{2}\mathbf{A}^2\mathbf{A}^{-1}\)	M1	3.1a
Identifies \(\mathbf{A}^{-1} = -\dfrac{1}{2}\mathbf{A} + \dfrac{5}{2}\mathbf{I}\)	A1	1.1b
	(3)
	(5 marks)

Notes

B1ft: Uses Cayley Hamilton Theorem to produce equation replacing \(\lambda\) with \(\mathbf{A}\) and constant term with constant multiple of the identity matrix \(\mathbf{I}\)

M1: A complete method using part (a) to find \(\mathbf{A}^{-1}\)
Multiplies through by \(\mathbf{A}^{-1}\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\)
Or rearranges to make \(\mathbf{I}\) the subject, takes out a factor of \(\mathbf{A}\), and rearranges to get \(\mathbf{A}^{-1} = \ldots\)
Or rearranges to make \(\mathbf{I}\) the subject and multiplies through by \(\mathbf{A}^{-1}\) to get \(\mathbf{A}^{-1} = \ldots\)

A1: Correct expression for \(\mathbf{A}^{-1}\), must be using their answer to part (a).