AS June 2020 Paper 1 Q16

AQACurrent spec4 marksMatrices

16 \(\mathbf{A}\) and \(\mathbf{B}\) are non-singular square matrices.

(a) Write down the product \(\mathbf{AA}^{-1}\) as a single matrix. [1 mark]

(b) \(\mathbf{M}\) is a matrix such that \(\mathbf{M} = \mathbf{AB}\).

Prove that \(\mathbf{M}^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) [3 marks]

Scheme	Marks	AO
Obtains \(\mathbf{I}\). Accept any identity matrix, eg \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).	B1	1.2

Mark scheme (b)

Scheme	Marks	AO
Correctly uses pre- or post - multiplication by either \(\mathbf{A}^{-1}\), \(\mathbf{B}^{-1}\) or \(\mathbf{M}^{-1}\) in a way which gives a product equal to \(\mathbf{I}\), starting from \(\mathbf{M} = \mathbf{AB}\) or \(\mathbf{ABM}^{-1} = \mathbf{I}\) or \(\mathbf{M}^{-1}\mathbf{AB} = \mathbf{I}\)	M1	1.1a
Correctly simplifies \(\mathbf{AA}^{-1} = \mathbf{I}\) OE or \(\mathbf{BB}^{-1} = \mathbf{I}\) OE in a correct equation.	M1	1.1a
Completes a rigorous argument to prove that \(\mathbf{M}^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\).	R1	2.1
	(4 marks)

Typical solution

\[\begin{aligned}\mathbf{M} &= \mathbf{AB} \\ \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{M} &= \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{AB} \\ \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{M} &= \mathbf{B}^{-1}\mathbf{IB} \\ \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{MM}^{-1} &= \mathbf{B}^{-1}\mathbf{BM}^{-1} \\ \mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{I} &= \mathbf{IM}^{-1} \\ \mathbf{B}^{-1}\mathbf{A}^{-1} &= \mathbf{M}^{-1}\end{aligned}\]