Question 16 - A-Level Maths

AQA Further AS Paper 1 2020 June — Question 16 4 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2020
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Matrices
Type	Properties of matrix operations
Difficulty	Moderate -0.8 Part (a) is trivial recall (AA^{-1} = I). Part (b) is a standard proof requiring verification that B^{-1}A^{-1} is the inverse of AB by showing their product equals I, using associativity of matrix multiplication. This is a textbook exercise in matrix algebra with no novel insight required, making it easier than average despite being Further Maths content.
Spec	4.03o Inverse 3x3 matrix 4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)

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

Write down the product \(\mathbf{AA}^{-1}\) as a single matrix. [1 mark]
\(\mathbf{M}\) is a matrix such that \(\mathbf{M} = \mathbf{AB}\). Prove that \(\mathbf{M}^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) [3 marks]

Show mark scheme Show mark scheme source

Question 16:

Answer	Marks
16(a)	Obtains .

Accept any identity matrix, eg .

𝐈𝐈

Answer	Marks	Guidance
1 0	1.2	B1

𝐀𝐀𝐀𝐀 = 𝐈𝐈

Answer	Marks
16(b)	� �

0 1

Correctly uses pre- or post - multiplication by either or in

a way which gives a product equal to , starting from −1 −1 −1

𝐀𝐀 ,𝐁𝐁 𝐌𝐌

𝐈𝐈

or

𝐌𝐌= 𝐀𝐀𝐁𝐁

or −1

𝐀𝐀𝐁𝐁𝐌𝐌 = 𝐈𝐈

−1

Answer	Marks	Guidance
𝐌𝐌 𝐀𝐀𝐁𝐁= 𝐈𝐈	1.1a	M1

−𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏

𝐁𝐁 𝐀𝐀 𝐌𝐌= 𝐁𝐁 𝐀𝐀 𝐀𝐀𝐁𝐁

−𝟏𝟏 −𝟏𝟏 −𝟏𝟏

𝐁𝐁 𝐀𝐀 𝐌𝐌= 𝐁𝐁 𝐈𝐈𝐁𝐁

−𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏

𝐁𝐁 𝐀𝐀 𝐌𝐌𝐌𝐌 = 𝐁𝐁 𝐁𝐁𝐌𝐌

−𝟏𝟏 −𝟏𝟏 −𝟏𝟏

𝐁𝐁 𝐀𝐀 𝐈𝐈= 𝐈𝐈𝐌𝐌

−𝟏𝟏 −𝟏𝟏 −𝟏𝟏

𝐁𝐁 𝐀𝐀 = 𝐌𝐌

Correctly simplifies OE

or OE −1

in a co−r𝟏𝟏rect equation𝐀𝐀. 𝐀𝐀 = 𝐈𝐈

Answer	Marks	Guidance
𝐁𝐁𝐁𝐁 = 𝐈𝐈	1.1a	M1

Completes a rigorous argument to prove that .

−1 −1 −1

Answer	Marks	Guidance
𝐌𝐌 = 𝐁𝐁 𝐀𝐀	2.1	R1
Total	4
Q	Marking instructions	AO

Question 16:
--- 16(a) ---
16(a) | Obtains .
Accept any identity matrix, eg .
𝐈𝐈
1 0 | 1.2 | B1 | −1
𝐀𝐀𝐀𝐀 = 𝐈𝐈
--- 16(b) ---
16(b) | � �
0 1
Correctly uses pre- or post - multiplication by either or in
a way which gives a product equal to , starting from −1 −1 −1
𝐀𝐀 ,𝐁𝐁 𝐌𝐌
𝐈𝐈
or
𝐌𝐌= 𝐀𝐀𝐁𝐁
or −1
𝐀𝐀𝐁𝐁𝐌𝐌 = 𝐈𝐈
−1
𝐌𝐌 𝐀𝐀𝐁𝐁= 𝐈𝐈 | 1.1a | M1 | 𝐌𝐌= 𝐀𝐀𝐁𝐁
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐁𝐁 𝐀𝐀 𝐌𝐌= 𝐁𝐁 𝐀𝐀 𝐀𝐀𝐁𝐁
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐁𝐁 𝐀𝐀 𝐌𝐌= 𝐁𝐁 𝐈𝐈𝐁𝐁
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐁𝐁 𝐀𝐀 𝐌𝐌𝐌𝐌 = 𝐁𝐁 𝐁𝐁𝐌𝐌
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐁𝐁 𝐀𝐀 𝐈𝐈= 𝐈𝐈𝐌𝐌
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐁𝐁 𝐀𝐀 = 𝐌𝐌
Correctly simplifies OE
or OE −1
in a co−r𝟏𝟏rect equation𝐀𝐀. 𝐀𝐀 = 𝐈𝐈
𝐁𝐁𝐁𝐁 = 𝐈𝐈 | 1.1a | M1
Completes a rigorous argument to prove that .
−1 −1 −1
𝐌𝐌 = 𝐁𝐁 𝐀𝐀 | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

$\mathbf{A}$ and $\mathbf{B}$ are non-singular square matrices.

\begin{enumerate}[label=(\alph*)]
\item Write down the product $\mathbf{AA}^{-1}$ as a single matrix.
[1 mark]

\item $\mathbf{M}$ is a matrix such that $\mathbf{M} = \mathbf{AB}$.

Prove that $\mathbf{M}^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q16 [4]}}

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