FP1 June 2018 Q3

EdexcelOld spec9 marksMatrices

(i) Given that \[\mathbf{A} = \begin{pmatrix} -2 & 3 \\ 1 & 1 \end{pmatrix}, \quad \mathbf{AB} = \begin{pmatrix} -1 & 5 & 12 \\ 3 & -5 & -1 \end{pmatrix}\]

(a) find \(\mathbf{A}^{-1}\) (2)

(b) Hence, or otherwise, find the matrix B, giving your answer in its simplest form. (3)

(ii) Given that \[\mathbf{C} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\]

(a) describe fully the single geometrical transformation represented by the matrix C. (2)

(b) Hence find the matrix \(\mathbf{C}^{39}\) (2)

Mark scheme (i)(a)

Scheme	Marks
\(\mathbf{A}^{-1} = \dfrac{1}{-2 - 3}\begin{pmatrix} 1 & -3 \\ -1 & -2 \end{pmatrix}\)	M1 A1
	(2)

Notes

M1: Either \(\dfrac{1}{-2 - 3}\) or \(-\dfrac{1}{5}\) or \(\begin{pmatrix} 1 & -3 \\ -1 & -2 \end{pmatrix}\)

A1: Correct expression for \(\mathbf{A}^{-1}\)

Mark scheme (i)(b)

Scheme	Marks
\(\left\{\mathbf{B} = \mathbf{A}^{-1}(\mathbf{AB})\right\}\)
\(\mathbf{B} = -\dfrac{1}{5}\begin{pmatrix} 1 & -3 \\ -1 & -2 \end{pmatrix}\begin{pmatrix} -1 & 5 & 12 \\ 3 & -5 & -1 \end{pmatrix}\)	M1
\(= \left\{-\dfrac{1}{5}\right\}\begin{pmatrix} -10 & 20 & 15 \\ -5 & 5 & -10 \end{pmatrix}\)	A1
\(= \begin{pmatrix} 2 & -4 & -3 \\ 1 & -1 & 2 \end{pmatrix}\)	A1
	(3)

Notes

M1: Writing down their \(\mathbf{A}^{-1}\) multiplied by AB

A1: At least one correct row or at least two correct columns of \(\begin{pmatrix} \ldots \\ \ldots \end{pmatrix}\). (Ignore \(-\dfrac{1}{5}\)).

A1: Correct simplified matrix for B

ALT (b)

Scheme	Marks
Let \(\mathbf{B} = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}\)
\(-2a + 3d = -1 \qquad -2b + 3e = 5\) \(a + d = 3 \qquad b + e = -5\) \(-2c + 3f = 12\) \(c + f = -1\)	M1
\(\{a = 2,\ d = 1,\ b = -4,\ e = -1,\ c = -3,\ f = 2\}\)
\(\mathbf{B} = \begin{pmatrix} 2 & -4 & -3 \\ 1 & -1 & 2 \end{pmatrix}\)	A1 A1
	(3)

M1: Writes down at least 2 correct sets of simultaneous equations

A1: At least one correct row or at least two correct columns for the matrix B

A1: Correct matrix for B

Mark scheme (ii)(a)

Scheme	Marks
Rotation	M1
\(90^\circ\) clockwise about the origin	A1
	(2)

Notes

M1: Rotation only.

A1: \(90^\circ\) \(\left(\text{or } \dfrac{\pi}{2}\right)\) clockwise about the origin
or \(270^\circ\) \(\left(\text{or } \dfrac{3\pi}{2}\right)\) (anti-clockwise) about the origin.
\(-90^\circ\) \(\left(\text{or } -\dfrac{\pi}{2}\right)\) (anticlockwise) about the origin. Origin can be written as \((0, 0)\) or O.

Mark scheme (ii)(b)

Scheme	Marks
\(\left\{\mathbf{C}^{39}\right\} = \mathbf{C}^{-1}\) or \(\mathbf{C}^3 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\)	M1 A1
	(2)
	(9 marks)

Notes

M1: For stating \(\mathbf{C}^{-1}\) or \(\mathbf{C}^3\) or ‘rotation of \(270^\circ\) clockwise o.e. about the origin. Can be implied by correct matrix.

A1: \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\). Correct answer with no working award M1A1