FP1 June 2017 Q5

EdexcelOld spec11 marksMatrices

(i) \[\mathbf{A} = \begin{pmatrix} p & 2 \\ 3 & p \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} -5 & 4 \\ 6 & -5 \end{pmatrix}\] where \(p\) is a constant.

(a) Find, in terms of \(p\), the matrix AB (2)

Given that \[\mathbf{AB} + 2\mathbf{A} = k\mathbf{I}\] where \(k\) is a constant and I is the \(2 \times 2\) identity matrix,

(b) find the value of \(p\) and the value of \(k\). (4)

(ii) \[\mathbf{M} = \begin{pmatrix} a & -9 \\ 1 & 2 \end{pmatrix}, \quad \text{where } a \text{ is a real constant}\] Triangle \(T\) has an area of 15 square units.
Triangle \(T\) is transformed to the triangle \(T^{\prime}\) by the transformation represented by the matrix M.
Given that the area of triangle \(T^{\prime}\) is 270 square units, find the possible values of \(a\). (5)

Mark scheme (i)(a)

Scheme	Marks
\(\mathbf{A} = \begin{pmatrix} p & 2 \\ 3 & p \end{pmatrix},\ \mathbf{B} = \begin{pmatrix} -5 & 4 \\ 6 & -5 \end{pmatrix},\ \mathbf{M} = \begin{pmatrix} a & -9 \\ 1 & 2 \end{pmatrix}\) \(p, a\) are constants.
\(\{\mathbf{AB}\} = \begin{pmatrix} -5p + 12 & 4p - 10 \\ -15 + 6p & 12 - 5p \end{pmatrix}\)	M1 A1
	(2)

Notes

M1: At least 2 elements are correct.

A1: Correct matrix.

Mark scheme (i)(b)

Scheme	Marks
\(\{\mathbf{AB} + 2\mathbf{A} = k\mathbf{I}\}\) \(\begin{pmatrix} -5p + 12 & 4p - 10 \\ -15 + 6p & 12 - 5p \end{pmatrix} + 2\begin{pmatrix} p & 2 \\ 3 & p \end{pmatrix} = k\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) \(\begin{pmatrix} -3p + 12 & 4p - 6 \\ -9 + 6p & 12 - 3p \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}\)
\(\text{"}4p - 10\text{"} + 4 = 0\) or \(\text{"}-15 + 6p\text{"} + 6 = 0\) or \(\text{"}-9 + 6p\text{"} = \text{"}4p - 6\text{"}\)	M1
\(\Rightarrow p = \dfrac{3}{2}\)	A1
\(k = -5\left(\dfrac{3}{2}\right) + 12 + 2\left(\dfrac{3}{2}\right) \Rightarrow k = \ldots\)	M1
\(k = \dfrac{15}{2}\)	A1
	(4)

Notes

If ‘simultaneous equations’ used, allocate marks as below.

M1: Forms an equation in \(p\)

A1: \(p = \dfrac{3}{2}\) o.e.

M1: Substitutes their \(p = \dfrac{3}{2}\) into "their \((-5p + 12)\)" \(+ 2p\) to find a value for \(k\) or eliminates \(p\) to find \(k\).

A1: \(k = \dfrac{15}{2}\) oe

Mark scheme (ii)

Scheme	Marks
Way 1 \(\pm\dfrac{270}{15}\ \{= \pm 18\}\)	B1
\(\det\mathbf{M} = (a)(2) - (-9)(1)\)	M1
\(\Rightarrow 2a + 9 = 18\) or \(2a + 9 = -18\)	M1
\(\Rightarrow a = 4.5\) or \(a = -13.5\)	A1 A1
	(5)
	(11 marks)

Notes

B1: Can be implied from calculations.

M1: Applies \(ad - bc\) to M. Require clear evidence of correct formula being used for M1 if errors seen.

M1: Equates their \(\det\mathbf{M}\) to either 18 or \(-18\) (corrected from the printed mark scheme: \(\det\mathbf{A}\))

A1: At least one of either \(a = 4.5\) or \(a = -13.5\)

A1: Both \(a = 4.5\) and \(a = -13.5\)

(ii) Way 2

Scheme	Marks
Consider vertices of triangle with area 15 units e.g. (0,0), (15,0) and (0,2) and attempting 2 values of \(a\).	B1
e.g. \(\begin{pmatrix} a & -9 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 0 & 15 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 15a & -18 \\ 0 & 15 & 4 \end{pmatrix}\)	M1
e.g. \(\dfrac{1}{2}\begin{vmatrix} 0 & 15a & -18 & 0 \\ 0 & 15 & 4 & 0 \end{vmatrix} = 270\)	M1
\(\Rightarrow a = 4.5\) or \(a = -13.5\)	A1 A1
	(5)

M1: Pre-multiplies their matrix by M and obtains single matrix

M1: Equates their determinant to 270 and attempts to solve.

A1: At least one of either \(a = 4.5\) or \(a = -13.5\)

A1: Both \(a = 4.5\) and \(a = -13.5\)