AS June 2022 Paper 1 Q1

EdexcelCurrent spec7 marksMatrices

\[\mathbf{A} = \begin{pmatrix}4 & -1\\ 7 & 2\\ -5 & 8\end{pmatrix}\qquad \mathbf{B} = \begin{pmatrix}2 & 3 & 2\\ -1 & 6 & 5\end{pmatrix}\qquad \mathbf{C} = \begin{pmatrix}-5 & 2 & 1\\ 4 & 3 & 8\\ -6 & 11 & 2\end{pmatrix}\]

Given that \(\mathbf{I}\) is the \(3 \times 3\) identity matrix,

(a)

(i) show that there is an integer \(k\) for which\[\mathbf{AB} - 3\mathbf{C} + k\mathbf{I} = \mathbf{0}\]stating the value of \(k\)

(ii) explain why there can be no constant \(m\) such that\[\mathbf{BA} - 3\mathbf{C} + m\mathbf{I} = \mathbf{0}\]

(4)

(b)

(i) Show how the matrix \(\mathbf{C}\) can be used to solve the simultaneous equations\[\begin{aligned}-5x + 2y + z &= -14\\ 4x + 3y + 8z &= 3\\ -6x + 11y + 2z &= 7\end{aligned}\]

(ii) Hence use your calculator to solve these equations.

(3)

Mark scheme (a)

Scheme	Marks	AO
(i)\[\mathbf{AB} = \begin{pmatrix}4 & -1\\ 7 & 2\\ -5 & 8\end{pmatrix}\begin{pmatrix}2 & 3 & 2\\ -1 & 6 & 5\end{pmatrix} = \begin{pmatrix}8 + 1 & 12 - 6 & 8 - 5\\ 14 - 2 & 21 + 12 & 14 + 10\\ -10 - 8 & -15 + 48 & -10 + 40\end{pmatrix} = \begin{pmatrix}9 & 6 & 3\\ 12 & 33 & 24\\ -18 & 33 & 30\end{pmatrix}\]	M1	1.1b
\[\text{So } \mathbf{AB} - 3\mathbf{C} = \begin{pmatrix}9 & 6 & 3\\ 12 & 33 & 24\\ -18 & 33 & 30\end{pmatrix} - \begin{pmatrix}-15 & 6 & 3\\ 12 & 9 & 24\\ -18 & 33 & 6\end{pmatrix} = \begin{pmatrix}24 & 0 & 0\\ 0 & 24 & 0\\ 0 & 0 & 24\end{pmatrix}\]or\[\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix}9 & 6 & 3\\ 12 & 33 & 24\\ -18 & 33 & 30\end{pmatrix} + \begin{pmatrix}15 & -6 & -3\\ -12 & -9 & -24\\ 18 & -33 & -6\end{pmatrix} = \begin{pmatrix}24 & 0 & 0\\ 0 & 24 & 0\\ 0 & 0 & 24\end{pmatrix}\]and states a value for \(k\)	M1	1.1b
Hence \(\mathbf{AB} - 3\mathbf{C} - 24\mathbf{I} = \mathbf{0}\) so \(k = -24\)	A1	1.1b
(ii) Need two things One of: BA is a 2×2 matrix Finds the matrix BA (must be a 2×2 matrix) AND One of: cannot subtract a 3×3 matrix finds matrix 3C and comments that they have different dimensions / can’t be done can’t subtract matrices of different sizes 3C or C is a 3×3 matrix BA needs to be a 3×3 matrix	B1	2.4
	(4)

Notes

(a)(i)
M1: Attempts to find AB. Usually this will be done on calculator so answer implies the method. If answer is incorrect allow for at least 6 correct entries or calculations shown.
This mark can be implied by a correct matrix for \(\mathbf{AB} - 3\mathbf{C}\) gives the first M1

M1: Uses their AB and 3C matrices to find a multiple I and states a value for \(k\)

A1: Correct proof with \(k = -24\) seen explicitly (may be in equation).

Minimum working required is \(\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix}24 & 0 & 0\\ 0 & 24 & 0\\ 0 & 0 & 24\end{pmatrix}\) gets M1 then states a value for \(k\) M1 then \(k = -24\) gets A1

Special case: If minimum working required is not seen and just \(k = -24\) stated then M1 M0 A0 as they have not shown that the value of \(k\) works.

(ii)
B1: Correct explanation referring to the dimensions of BA and C (or 3C) and that they do not match in the equation. They can find both these matrices and then comment they cannot be subtracted.

Mark scheme (b)

Scheme	Marks	AO
(i)\[\begin{pmatrix}-5 & 2 & 1\\ 4 & 3 & 8\\ -6 & 11 & 2\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}-14\\ 3\\ 7\end{pmatrix}\Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}-5 & 2 & 1\\ 4 & 3 & 8\\ -6 & 11 & 2\end{pmatrix}^{-1}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix}\]Or states \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \mathbf{C}^{-1}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix}\) Or states \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \dfrac{1}{360}\begin{pmatrix}-82 & 7 & 13\\ -56 & -4 & 44\\ 62 & 43 & -23\end{pmatrix}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix}\)	M1	1.2
(ii)\[= \dfrac{1}{360}\begin{pmatrix}-82 & 7 & 13\\ -56 & -4 & 44\\ 62 & 43 & -23\end{pmatrix}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix} = \ldots\]\[= \begin{pmatrix}-\dfrac{41}{180} & \dfrac{7}{360} & \dfrac{13}{360}\\[8pt] -\dfrac{7}{45} & -\dfrac{1}{90} & \dfrac{11}{90}\\[8pt] \dfrac{31}{180} & \dfrac{43}{360} & -\dfrac{23}{360}\end{pmatrix}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix} = \ldots\]\(\mathbf{C}^{-1}\begin{pmatrix}-14\\ 3\\ 7\end{pmatrix} = \ldots\)	M1	1.1b
So solution is \(x = \dfrac{7}{2},\ y = 3,\ z = -\dfrac{5}{2}\) or \((3.5,\ 3,\ -2.5)\)	A1	1.1b
	(3)
	(7 marks)

Notes

(b) Mark (i) and (ii) altogether

M1: States or implies use of the correct method of using the inverse matrix.

M1: Carries out the process of multiplying after finding the inverse. May find inverse long hand first. Finding the inverse matrix then writes down an answer gains M1.

Note: There is no need to find the inverse matrix. If the inverse matrix is not stated just answers written down then two out of the three correct ordinates imply the M1.

A1: Correct solution. Must be clear that \(x = \dfrac{7}{2},\ y = 3,\ z = -\dfrac{5}{2}\) allow \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}3.5\\ 3\\ -2.5\end{pmatrix}\)

Note: If they solve using simultaneous equations only this is M0 M0 A0
If there is no reference to the inverse matrix and correct answers stated this is M0 M0 A0