AS June 2024 Paper 1 Q4

EdexcelCurrent spec8 marksMatrices

\[\mathbf{A} = \begin{pmatrix}-1 & -2 & -7\\ 3 & k & 2\\ 1 & 1 & 4\end{pmatrix}\qquad \mathbf{B} = \begin{pmatrix}4k - 2 & 1 & 7k - 4\\ -10 & 3 & -19\\ 3 - k & -1 & 6 - k\end{pmatrix}\]

where \(k\) is a constant.

(a) Determine the value of the constant \(c\) for which\[\mathbf{AB} = (3k + c)\mathbf{I}\] (2)

(b) Hence determine the value of \(k\) for which \(\mathbf{A}^{-1}\) does not exist. (2)

Given that \(\mathbf{A}^{-1}\) does exist,

(d) Use the answer to part (c) to solve the simultaneous equations\[\begin{aligned}-x - 2y - 7z &= 10\\ 3x + ky + 2z &= 3\\ x + y + 4z &= 1\end{aligned}\]giving the values of \(x\), \(y\) and \(z\) in simplest form in terms of \(k\). (3)

Mark scheme (a)

Scheme	Marks	AO
\(-4k + 2 + 20 - 21 + 7k\) or \(3 + 3k - 2\) or \(7k - 4 - 19 + 24 - 4k\) or \(3k + 1\)	M1	1.1b
\(\{1 + 3k = 3k + c\}\) \(\Rightarrow c = 1\)	A1	1.1b
	(2)

Notes

M1: Calculates one of the elements of the leading diagonal of AB, condone sign slips

A1: Sets diagonal = \(3k + c\) and deduces the correct value for \(c\). Award for sight of \(3k + 1\)

Mark scheme (b)

Scheme	Marks	AO
\(3k + 1 = 0 \Rightarrow k = \ldots\) Or Attempts the determinant and sets = 0 leading to a value for \(k\)	M1	1.1b
\(\Rightarrow k = -\dfrac{1}{3}\)	A1ft	1.1b
	(2)

Notes

M1: Attempts to solve \(3k + \text{“}1\text{”} = 0\) or attempts the determinant, condone sign slips in the minors, and sets = 0 leading to a value for \(k\)

A1ft: Correct value or follow through their value for \(c\) so allow for \(k = -\dfrac{c}{3}\)

Mark scheme (c)

Scheme	Marks	AO
\(\left\{\mathbf{A}^{-1}\right\} = \dfrac{1}{3k + 1}\begin{pmatrix}4k - 2 & 1 & 7k - 4\\ -10 & 3 & -19\\ 3 - k & -1 & 6 - k\end{pmatrix}\)	B1ft	2.2a
	(1)

Notes

B1ft: Deduces the correct inverse matrix. Follow through their \(c\) so allow for \(\dfrac{1}{3k + c}\mathbf{B}\) or if found determinant \(\dfrac{1}{\text{their det}}\mathbf{B}\)

Mark scheme (d)

Scheme	Marks	AO
\(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \dfrac{1}{3k + 1}\begin{pmatrix}4k - 2 & 1 & 7k - 4\\ -10 & 3 & -19\\ 3 - k & -1 & 6 - k\end{pmatrix}\begin{pmatrix}10\\ 3\\ 1\end{pmatrix} = \ldots\)	M1	1.2
\(\left(\dfrac{47k - 21}{3k + 1},\ -\dfrac{110}{3k + 1},\ \dfrac{33 - 11k}{3k + 1}\right)\)	A1 A1	1.1b 1.1b
	(3)
	(8 marks)

Notes

M1: Complete method to find the values of \(x\), \(y\) and \(z\) using their inverse matrix

A1: At least one correct coordinate simplified or unsimplified.

A1: All coordinates correct and simplified. Condone as a column vector. Does not need to be written as a coordinate.

SC If candidate writes \(\dfrac{1}{3k + 1}\begin{pmatrix}4k - 2 & 1 & 7k - 4\\ -10 & 3 & -19\\ 3 - k & -1 & 6 - k\end{pmatrix}\begin{pmatrix}10\\ 3\\ 1\end{pmatrix}\) but ends up with at least one of \(x = 47k - 21,\ y = -110,\ z = 33 - 11k\) scores M1 A1 A0