A2 June 2025 Paper 2 Q3

OCR ACurrent spec9 marksMatrices

3 Matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = \begin{pmatrix} 1 & -3 \\ 4 & 8 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} -3 & -3 \\ 1 & 2 \end{pmatrix}\).

(a) Find the matrix \(\mathbf{AB}\). [1]

(b) Verify that \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\). [2]

The transformation represented by matrix \(\mathbf{A}\) is denoted by T.

(d) Show that the point \((2, -5)\) is not an invariant point under T. [2]

(e) Find the matrix which represents the inverse transformation of T. [2]

Scheme	Marks	AO
\((\mathbf{AB} =)\ \begin{pmatrix} -6 & -9 \\ -4 & 4 \end{pmatrix}\)	B1	1.1
	[1]

Mark scheme (b)

Scheme	Marks	AO
\(\det(\mathbf{AB}) = -60\) or \(\det\mathbf{A} = 20\) or \(\det\mathbf{B} = -3\)	M1	1.1
\(\det(\mathbf{A}) \times \det(\mathbf{B}) = 20 \times (-3)\) \(= -60 = \det(\mathbf{AB})\)	A1	2.1
	[2]

Notes

M1: Correctly finds at least one relevant determinant (including for their ‘AB’).

A1: Complete, connected and convincing argument, including all 3 determinants correct and multiplication giving \(-60\).
Allow \(20 \times (-3) = -60\) or \(\det(\mathbf{A}) \times \det(\mathbf{B}) = -60\) (provided determinants found first)

Mark scheme (c)

Scheme	Marks	AO
\(\mathbf{BA} = \begin{pmatrix} -15 & -15 \\ 9 & 13 \end{pmatrix}\)	B1*	1.1
\(\mathbf{BA} \neq \mathbf{AB}\)	B1dep*	2.1
	[2]

Notes

B1*: Finds BA.
Or finds one element and explicitly compares to corresponding element in AB.

B1dep*: Clear concluding statement following correct work.

Mark scheme (d)

Scheme	Marks	AO
\(\begin{pmatrix} 1 & -3 \\ 4 & 8 \end{pmatrix}\begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} 17 \\ -32 \end{pmatrix}\)	M1	1.1
\(\neq \begin{pmatrix} 2 \\ -5 \end{pmatrix}\) (so \((2, -5)\) is not an invariant point)	A1	2.4
	[2]

Notes

M1: Attempt to multiply the vector into the matrix, one element correct

A1: For clear conclusion from correct working
Values shown must be correct. Could be explained in words.

Mark scheme (e)

Scheme	Marks	AO
\(\mathbf{A}^{-1}\) soi as answer	M1	1.2
\(\frac{1}{20}\begin{pmatrix} 8 & 3 \\ -4 & 1 \end{pmatrix}\)	A1FT	2.2a
	[2]

Notes

M1: Recalling the connection between inverse transformation and inverse matrix.
Could be stated in words or implied by an attempt at the inverse e.g. \(A^{-1} = \cdots\) or \(\frac{1}{n}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\) etc.

A1FT: FT their det A from (b) if not recalculated. ISW.
\(\begin{pmatrix} \frac{2}{5} & \frac{3}{20} \\ -\frac{1}{5} & \frac{1}{20} \end{pmatrix}\) or \(\begin{pmatrix} 0.4 & 0.15 \\ -0.2 & 0.05 \end{pmatrix}\)