AS June 2024 Paper 1 Q11

AQACurrent spec3 marksComplex Numbers Matrices

11 The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by

\[\mathbf{A} = \begin{bmatrix} 3\mathrm{i} & -2 \\ a & -\mathrm{i} \end{bmatrix} \qquad \text{and} \qquad \mathbf{B} = \begin{bmatrix} 4 & 5 \\ -2\mathrm{i} & -1 \end{bmatrix}\]

where \(a\) is a real number.

Calculate the product \(\mathbf{AB}\) in terms of \(a\)

Give your answer in its simplest form. [3 marks]

Mark scheme

Scheme	Marks	AO
Correct method for at least two elements of \(\mathbf{AB}\)	M1	1.1a
Obtains \(4a - 2\) in row 2 column 1 of \(\mathbf{AB}\)	B1	1.1b
Obtains \(\begin{bmatrix} 16\mathrm{i} & 2 + 15\mathrm{i} \\ 4a - 2 & 5a + \mathrm{i} \end{bmatrix}\)	A1	1.1b
	(3 marks)

Typical solution

\[\begin{aligned}\mathbf{AB} &= \begin{bmatrix} 3\mathrm{i} & -2 \\ a & -\mathrm{i} \end{bmatrix}\begin{bmatrix} 4 & 5 \\ -2\mathrm{i} & -1 \end{bmatrix} \\ &= \begin{bmatrix} 12\mathrm{i} + 4\mathrm{i} & 15\mathrm{i} + 2 \\ 4a + 2\mathrm{i}^2 & 5a + \mathrm{i} \end{bmatrix} \\ &= \begin{bmatrix} 16\mathrm{i} & 2 + 15\mathrm{i} \\ 4a - 2 & 5a + \mathrm{i} \end{bmatrix}\end{aligned}\]