FP1 June 2013 Q8

EdexcelOld spec8 marksMatrices

8. \[\mathbf{A} = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\]

and \(\mathbf{I}\) is the \(2 \times 2\) identity matrix.

(a) Prove that \[\mathbf{A}^2 = 7\mathbf{A} + 2\mathbf{I}\] (2)

(b) Hence show that \[\mathbf{A}^{-1} = \frac{1}{2}(\mathbf{A} - 7\mathbf{I})\] (2)

The transformation represented by \(\mathbf{A}\) maps the point \(P\) onto the point \(Q\).

Given that \(Q\) has coordinates \((2k + 8,\ -2k - 5)\), where \(k\) is a constant,

Mark scheme (a)

Scheme	Marks
\(\mathbf{A}^2 = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} 44 & -14 \\ -28 & 9 \end{pmatrix}\) \(7\mathbf{A} + 2\mathbf{I} = \begin{pmatrix} 42 & -14 \\ -28 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 44 & -14 \\ -28 & 9 \end{pmatrix}\) M1: Attempt both \(\mathbf{A}^2\) and \(7\mathbf{A} + 2\mathbf{I}\) A1: Both matrices correct	M1A1
OR \(\mathbf{A}^2 - 7\mathbf{A} = \mathbf{A}(\mathbf{A} - 7\mathbf{I})\) M1 for expression and attempt to substitute and multiply (2x2)(2x2)=2x2
\(\mathbf{A}(\mathbf{A} - 7\mathbf{I}) = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2\mathbf{I}\) A1 cso
	(2)

Mark scheme (b)

Scheme	Marks
\(\mathbf{A}^2 = 7\mathbf{A} + 2\mathbf{I} \Rightarrow \mathbf{A} = 7\mathbf{I} + 2\mathbf{A}^{-1}\) Require one correct line using accurate expressions involving \(\mathbf{A}^{-1}\) and identity matrix to be clearly stated as \(\mathbf{I}\).	M1
\(\mathbf{A}^{-1} = \dfrac{1}{2}(\mathbf{A} - 7\mathbf{I})\) *	A1* cso
	(2)

Numerical approach award 0/2.

Mark scheme (c)

Scheme	Marks
\(\mathbf{A}^{-1} = \dfrac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}\) Correct inverse matrix or equivalent	B1
\(\dfrac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}\begin{pmatrix} 2k + 8 \\ -2k - 5 \end{pmatrix} = \dfrac{1}{2}\begin{pmatrix} -2k - 8 + 4k + 10 \\ -8k - 32 + 12k + 30 \end{pmatrix}\) Matrix multiplication involving their inverse and \(k\): (2x2)(2x1)=2x1. N.B. \(\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 2k + 8 \\ -2k - 5 \end{pmatrix}\) is M0	M1
\(\begin{pmatrix} k + 1 \\ 2k - 1 \end{pmatrix}\) or \((k + 1,\ 2k - 1)\) \((k + 1)\) first A1, \((2k - 1)\) second A1	A1,A1
	(4)
	Total 8

Scheme	Marks
\(\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2k + 8 \\ -2k - 5 \end{pmatrix}\) Correct matrix equation.	B1
\(6x - 2y = 2k + 8\) \(-4x + y = -2k - 5 \Rightarrow x = \ldots\) or \(y = \ldots\) Multiply out and attempt to solve simultaneous equations for \(x\) or \(y\) in terms of \(k\).	M1
\(\begin{pmatrix} k + 1 \\ 2k - 1 \end{pmatrix}\) or \((k + 1,\ 2k - 1)\) \((k + 1)\) first A1, \((2k - 1)\) second A1	A1,A1
	(4)