A2 October 2021 Paper 1 Q4

EdexcelCurrent spec9 marksMatrices

(i) \(\mathbf{A}\) is a 2 by 2 matrix and \(\mathbf{B}\) is a 2 by 3 matrix.
Giving a reason for your answer, explain whether it is possible to evaluate

(a) \(\mathbf{AB}\)

(b) \(\mathbf{A} + \mathbf{B}\) (2)

(ii) Given that\[\begin{pmatrix}-5 & 3 & 1\\ a & 0 & 0\\ b & a & b\end{pmatrix}\begin{pmatrix}0 & 5 & 0\\ 2 & 12 & -1\\ -1 & -11 & 3\end{pmatrix} = \lambda\mathbf{I}\]where \(a\), \(b\) and \(\lambda\) are constants,

(a) determine

the value of \(\lambda\)
the value of \(a\)
the value of \(b\)

(b) Hence deduce the inverse of the matrix \(\begin{pmatrix}-5 & 3 & 1\\ a & 0 & 0\\ b & a & b\end{pmatrix}\) (3)

(iii) Given that\[\mathbf{M} = \begin{pmatrix}1 & 1 & 1\\ 0 & \sin\theta & \cos\theta\\ 0 & \cos 2\theta & \sin 2\theta\end{pmatrix} \qquad \text{where } 0 \leqslant \theta \lt \pi\]determine the values of \(\theta\) for which the matrix \(\mathbf{M}\) is singular. (4)

Mark scheme (i)

Scheme	Marks	AO
(a) It is possible as the number of columns of matrix A matches the number of rows of matrix B.	B1	2.4
(b) It is not possible as matrix A and matrix B have different dimensions o.e. different number of columns	B1	2.4
	(2)

Notes

(a) B1: Comments that the number of columns of matrix A (2) equals the number of rows of matrix B (2) therefore it is possible. Accept other terminology that is clear in intent e.g. “length of A” and “height of B”

(b) B1: Comments that matrix A and matrix B have different dimensions therefore it is not possible.

Mark scheme (ii)

Scheme	Marks	AO
(a) \(\lambda = 5\)	B1	2.2a
\(a = 1,\ b = 2\)	B1	2.2a
(b) Inverse matrix \(= \dfrac{1}{5}\begin{pmatrix}0 & 5 & 0\\ 2 & 12 & -1\\ -1 & -11 & 3\end{pmatrix}\)	B1ft	3.1a
	(3)

Notes

(a) B1: Deduces the correct value for \(\lambda = 5\)

B1: Deduces the correct values for \(a\) and \(b\)

(b) B1ft: Identifies and applies a correct method find the inverse matrix. May multiply from the given equation, in which case follow through on their value of lambda. Alternatively, award for a correct matrix found by calculator or long hand having found \(a\) and \(b\) and using these values in the matrix.

Mark scheme (iii)

Scheme	Marks	AO
A complete method to find the determinant of the matrix and set equal to zero.	M1	1.1b
Determinant \(= 1(\sin\theta\sin 2\theta - \cos\theta\cos 2\theta) - 1(0) + 1(0) = 0\)	A1	1.1b
Uses compound angle formula to achieve \(\cos 3\theta = 0\) leading to \(\theta = \ldots\) or use of \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 1 - 2\sin^2\theta\) (e.g. to achieve \(\cos\theta\left(4\sin^2\theta - 1\right) = 0\)) leading to \(\theta = \ldots\) or use of \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 2\cos^2\theta - 1\) (e.g. to achieve \(4\cos^3\theta - 3\cos\theta = 0\)) leading to \(\theta = \ldots\)	M1	3.1a
\(\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}\)	A1	1.1b
	(4)
	(9 marks)

Notes

M1: A complete method to find the determinant of the matrix and sets it equal to 0

A1: Correct equation

M1: Uses appropriate correct trig identities to solve the equation and finds a value for \(\theta\)

A1: All three correct values \(\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}\) and no others in the range.