FP1 June 2014 (R) Q6

EdexcelOld spec10 marksMatrices

6. \[\mathbf{A} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} \text{ and } \mathbf{B} = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}\]

Given that \(\mathbf{M} = (\mathbf{A} + \mathbf{B})(2\mathbf{A} - \mathbf{B})\),

(a) calculate the matrix \(\mathbf{M}\), (6)

(b) find the matrix \(\mathbf{C}\) such that \(\mathbf{MC} = \mathbf{A}\). (4)

Mark scheme (a)

Scheme	Marks
\(\mathbf{A} + \mathbf{B} = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\) M1: Correct attempt at matrix addition with 3 elements correct A1: Correct matrix	M1A1
\(2\mathbf{A} - \mathbf{B} = \begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix}\) M1: Correct attempt to double \(\mathbf{A}\) and subtract \(\mathbf{B}\) 3 elements correct A1: Correct matrix	M1A1
\((\mathbf{A} + \mathbf{B})(2\mathbf{A} - \mathbf{B}) = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\) M1: Correct method to multiply A1: cao	M1A1
	(6)

Scheme	Marks
\((\mathbf{A} + \mathbf{B})(2\mathbf{A} - \mathbf{B}) = 2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2\) M1: Expands brackets with at least 3 correct terms A1: Correct expansion	M1A1
\(\mathbf{A}^2 = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}, \mathbf{BA} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix},\) \(\mathbf{AB} = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix}, \mathbf{B}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) M1: Attempts \(\mathbf{A}^2\), \(\mathbf{B}^2\) and \(\mathbf{AB}\) or \(\mathbf{BA}\) A1: Correct matrices	M1A1
\(2\mathbf{A}^2 + 2\mathbf{BA} - \mathbf{AB} - \mathbf{B}^2 = \begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\) M1: Substitutes into their expansion A1: Correct matrix	M1A1

Mark scheme (b)

Scheme	Marks
\(\mathbf{MC} = \mathbf{A} \Rightarrow \mathbf{C} = \mathbf{M}^{-1}\mathbf{A}\) May be implied by later work	B1
\(\mathbf{M}^{-1} = \dfrac{1}{-2 - 7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\) An attempt at their \(\dfrac{1}{\det\mathbf{M}}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\)	M1
\(\mathbf{C} = \dfrac{1}{-2 - 7}\begin{pmatrix} -2 & 1 \\ 7 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\) Correct order required and an attempt to multiply	dM1
\(\mathbf{C} = -\dfrac{1}{9}\begin{pmatrix} -5 & -2 \\ 13 & 7 \end{pmatrix}\) oe	A1
	(4)
	(10 marks)

Scheme	Marks
\(\begin{pmatrix} 1 & -1 \\ -7 & -2 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\) Correct statement	B1
\(a - c = 2,\ b - d = 1\) \(-7a - 2c = -1,\ -7b - 2d = 0\) Multiplies correctly to obtain 4 equations	M1
\(a = \dfrac{5}{9}, b = \dfrac{2}{9}, c = -\dfrac{13}{9}, d = -\dfrac{7}{9}\) M1: Solves to obtain values for \(a\), \(b\), \(c\) and \(d\) A1: Correct values	M1A1