Question 16 - A-Level Maths

AQA Further AS Paper 1 2018 June — Question 16 3 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2018
Session	June
Marks	3
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Matrices
Type	Solving matrix equations for unknown matrix
Difficulty	Standard +0.8 This is a Further Maths matrix equation requiring algebraic manipulation to isolate A. Students must rearrange AB = I + 2A to A(B - 2I) = I, then find the inverse of (B - 2I). While the steps are systematic, this requires understanding matrix algebra beyond standard A-level and careful computation of a 2×2 inverse, making it moderately challenging but still a standard Further Maths exercise.
Spec	4.03a Matrix language: terminology and notation 4.03b Matrix operations: addition, multiplication, scalar 4.03o Inverse 3x3 matrix

Two matrices $\mathbf{A}$ and $\mathbf{B}$ satisfy the equation $$\mathbf{AB} = \mathbf{I} + 2\mathbf{A}$$ where $\mathbf{I}$ is the identity matrix and $\mathbf{B} = \begin{pmatrix} 3 & -2 \\ -4 & 8 \end{pmatrix}$ Find $\mathbf{A}$. [3 marks]

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Question 16:

Answer	Marks	Guidance
16	Uses factorisation or pre-multiplication to isolate
𝑨𝑨	3.1a	M1

𝑨𝑨(𝑨𝑨−2𝑰𝑰)= 𝑰𝑰

−1

𝑨𝑨 = (𝑨𝑨−2𝑰𝑰)

−1

1 −2

𝑨𝑨 = � �

−4 6

1 6 2

𝑨𝑨 = −2� �

4 1

−3 −1

Deduces in terms of and .

Could be implied by sight of with attempt to invert.

𝑨𝑨 𝑨𝑨 𝑰𝑰

Answer	Marks	Guidance
1 −2	2.2a	A1

� �

−4 6

Obtains correct matrix .

Answer	Marks	Guidance
𝑨𝑨	1.1b	A1

ALT

Answer	Marks	Guidance
16	Sets up four equations with at least three correct.	3.1a

𝑎𝑎 𝑏𝑏

𝑨𝑨 = � �

𝑐𝑐 𝑑𝑑

𝑎𝑎 𝑏𝑏 3 −2 1 0 𝑎𝑎 𝑏𝑏

� �� = � �+2� �

and

𝑐𝑐 𝑑𝑑 −4 8 0 1 𝑐𝑐 𝑑𝑑

a3n𝑎𝑎d− 4𝑏𝑏 = 1+2𝑎𝑎 3 𝑐𝑐−4𝑑𝑑 = 0+2𝑐𝑐

and −2𝑎𝑎+8𝑏𝑏 = 0+2𝑏𝑏

−2𝑐𝑐+8𝑑𝑑 = 1+2𝑑𝑑

and

𝑎𝑎 = 4𝑏𝑏+1 𝑐𝑐 = 4𝑑𝑑

and

6𝑏𝑏 = 2𝑎𝑎 6𝑑𝑑 = 2𝑐𝑐+1

and

∴ 3𝑏𝑏 = 4𝑏𝑏+1 6𝑑𝑑 = 2(4𝑑𝑑)+1

1 −1 = 𝑏𝑏 −1 = a2n𝑑𝑑d ⇒ 𝑑𝑑 = −2

1 and

∴ 𝑎𝑎 = 4×−1+1 𝑐𝑐 = 4×−2

𝑎𝑎 = −3 𝑐𝑐 = −2

−3 −1

Deduces at least two correct elements of .

Note: Two correct elements from just two correct equations can score M1A1.

Answer	Marks	Guidance
𝑨𝑨	2.2a	A1
Obtains correct matrix	1.1b	A1

𝑨𝑨

Answer	Marks	Guidance
Total	3	∴ 𝑨𝑨 = � �

−2 −0.5

Answer	Marks	Guidance
Q	Marking instructions	AO

Question 16:
16 | Uses factorisation or pre-multiplication to isolate
𝑨𝑨 | 3.1a | M1 | 𝑨𝑨𝑨𝑨−2𝑨𝑨 = 𝑰𝑰
𝑨𝑨(𝑨𝑨−2𝑰𝑰)= 𝑰𝑰
−1
𝑨𝑨 = (𝑨𝑨−2𝑰𝑰)
−1
1 −2
𝑨𝑨 = � �
−4 6
1
6 2
𝑨𝑨 = −2� �
4 1
−3 −1
Deduces in terms of and .
Could be implied by sight of with attempt to invert.
𝑨𝑨 𝑨𝑨 𝑰𝑰
1 −2 | 2.2a | A1
� �
−4 6
Obtains correct matrix .
𝑨𝑨 | 1.1b | A1
ALT
16 | Sets up four equations with at least three correct. | 3.1a | M1 | Let
𝑎𝑎 𝑏𝑏
𝑨𝑨 = � �
𝑐𝑐 𝑑𝑑
𝑎𝑎 𝑏𝑏 3 −2 1 0 𝑎𝑎 𝑏𝑏
� �� �= � �+2� �
and
𝑐𝑐 𝑑𝑑 −4 8 0 1 𝑐𝑐 𝑑𝑑
a3n𝑎𝑎d− 4𝑏𝑏 = 1+2𝑎𝑎 3 𝑐𝑐−4𝑑𝑑 = 0+2𝑐𝑐
and −2𝑎𝑎+8𝑏𝑏 = 0+2𝑏𝑏
−2𝑐𝑐+8𝑑𝑑 = 1+2𝑑𝑑
and
and
𝑎𝑎 = 4𝑏𝑏+1 𝑐𝑐 = 4𝑑𝑑
and
6𝑏𝑏 = 2𝑎𝑎 6𝑑𝑑 = 2𝑐𝑐+1
and
∴ 3𝑏𝑏 = 4𝑏𝑏+1 6𝑑𝑑 = 2(4𝑑𝑑)+1
1
−1 = 𝑏𝑏 −1 = a2n𝑑𝑑d ⇒ 𝑑𝑑 = −2
1
and
∴ 𝑎𝑎 = 4×−1+1 𝑐𝑐 = 4×−2
𝑎𝑎 = −3 𝑐𝑐 = −2
−3 −1
Deduces at least two correct elements of .
Note: Two correct elements from just two correct equations can score M1A1.
𝑨𝑨 | 2.2a | A1
Obtains correct matrix | 1.1b | A1
𝑨𝑨
Total | 3 | ∴ 𝑨𝑨 = � �
−2 −0.5
Q | Marking instructions | AO | Mark | Typical solution

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Two matrices $\mathbf{A}$ and $\mathbf{B}$ satisfy the equation
$$\mathbf{AB} = \mathbf{I} + 2\mathbf{A}$$
where $\mathbf{I}$ is the identity matrix and $\mathbf{B} = \begin{pmatrix} 3 & -2 \\ -4 & 8 \end{pmatrix}$

Find $\mathbf{A}$.
[3 marks]

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q16 [3]}}

This paper (19 questions)

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Q1 1 Q2 1 Q3 1 Q4 2 Q5 3 Q6 3 Q7 2 Q8 5 Q9 6 Q10 8 Q11 3 Q12 6 Q13 9 Q14 7 Q15 4 Q16 3 Q17 4 Q18 4 Q19 8

AQA Further AS Paper 1 2018 June — Question 16 3 marks

Question 16:

𝑨𝑨(𝑨𝑨−2𝑰𝑰)= 𝑰𝑰

−1

𝑨𝑨 = (𝑨𝑨−2𝑰𝑰)

−1

1 −2

𝑨𝑨 = � �

−4 6

1

6 2

𝑨𝑨 = −2� �

4 1

−3 −1

Deduces in terms of and .

Could be implied by sight of with attempt to invert.

𝑨𝑨 𝑨𝑨 𝑰𝑰

� �

−4 6

Obtains correct matrix .

ALT

𝑎𝑎 𝑏𝑏

𝑨𝑨 = � �

𝑐𝑐 𝑑𝑑

𝑎𝑎 𝑏𝑏 3 −2 1 0 𝑎𝑎 𝑏𝑏

� �� �= � �+2� �

and

𝑐𝑐 𝑑𝑑 −4 8 0 1 𝑐𝑐 𝑑𝑑

a3n𝑎𝑎d− 4𝑏𝑏 = 1+2𝑎𝑎 3 𝑐𝑐−4𝑑𝑑 = 0+2𝑐𝑐

and −2𝑎𝑎+8𝑏𝑏 = 0+2𝑏𝑏

−2𝑐𝑐+8𝑑𝑑 = 1+2𝑑𝑑

and

and

𝑎𝑎 = 4𝑏𝑏+1 𝑐𝑐 = 4𝑑𝑑

and

6𝑏𝑏 = 2𝑎𝑎 6𝑑𝑑 = 2𝑐𝑐+1

and

∴ 3𝑏𝑏 = 4𝑏𝑏+1 6𝑑𝑑 = 2(4𝑑𝑑)+1

1

−1 = 𝑏𝑏 −1 = a2n𝑑𝑑d ⇒ 𝑑𝑑 = −2

1

and

∴ 𝑎𝑎 = 4×−1+1 𝑐𝑐 = 4×−2

𝑎𝑎 = −3 𝑐𝑐 = −2

−3 −1

Deduces at least two correct elements of .

Note: Two correct elements from just two correct equations can score M1A1.

𝑨𝑨

−2 −0.5

This paper (19 questions)

� �� = � �+2� �