AQA Further Paper 2 2024 June — Question 14 10 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind inverse then solve system
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring matrix inversion (using cofactors/adjugate method with a parameter), determining singularity conditions via determinant, and proving parameter-independence of solutions. While the techniques are standard Further Maths content, the algebraic manipulation with parameter k and the conceptual insight needed in part (c) to show independence elevates this above routine exercises.
Spec4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix
The matrix \(\mathbf{M}\) is defined as $$\mathbf{M} = \begin{bmatrix} 5 & 2 & 1 \\ 6 & 3 & 2k + 3 \\ 2 & 1 & 5 \end{bmatrix}$$ where \(k\) is a constant.
  1. Given that \(\mathbf{M}\) is a non-singular matrix, find \(\mathbf{M}^{-1}\) in terms of \(k\) [5 marks]
  2. State any restrictions on the value of \(k\) [1 mark]
  3. Using your answer to part (a), show that the solution to the set of simultaneous equations below is independent of the value of \(k\) \(5x + 2y + z = 1\) \(6x + 3y + (2k + 3)z = 4k + 3\) \(2x + y + 5z = 9\) [4 marks]
Question 14:
AnswerMarks Guidance
14(a)Obtains 12 – 2k 1.1b
= 12 – 2k
Cofactors are
–2k+12 4k –24 0 
 
–9 23 –1
 
  4k+3 –10k –9 3  
–2k+12 –9 4k+3 
1  
M –1 = 4k –24 23 –10k –9
 
12–2k
  0 −1 3  
Obtains matrix of
minors/cofactors with
at least four correct
elements.
PI by transposed
form.
Condone overall sign
error on each
AnswerMarks Guidance
element.1.1a M1
Obtains matrix of
minors/cofactors with
at least seven correct
elements.
PI transposed form.
Condone overall sign
error on each
AnswerMarks Guidance
element.1.1a M1
Obtains correct matrix
of minors/cofactors.
PI transposed form.
Condone overall sign
AnswerMarks Guidance
error on one element.1.1a M1
Obtains fully correct,
AnswerMarks Guidance
simplified answer.1.1b A1
Subtotal5
QMarking Instructions AO
AnswerMarks
14(b)Obtains k ≠ 6
Follow through their
AnswerMarks Guidance
determinant.1.1b B1F
Subtotal1
QMarking instructions AO
AnswerMarks
14(c)Uses their M –1 to form
a product to find the
solution set.
Must include
 1 
 
4k+3
 
  9  
or
Obtains
 1 
M –1 4k+3 
 
  9  
for a particular value
AnswerMarks Guidance
of k3.1a M1
  1   
y = 4k–24 23 –10k–9 4k+3
    
12–2k
 z    0 −1 3     9  
–2k+12–36k–27+36k+27
1  
= 4k–24+92k+69–90k–81
 
12–2k
  –4k–3+27  
 –2k+12
1  
= –36+6k
 
12–2k
 24–4k
 
 1 
 
= –3
 
  2  
which is independent of k
Obtains at least one
correct component
from their M –1, can be
unsimplified.
or
Obtains correct
 1 
–1  
M 4k+3
 
  9  
AnswerMarks Guidance
for their k1.1b A1F
Obtains at least two
correct components
from their M –1 (can be
unsimplified).
or
Correctly substitutes
their
 1 
–1  
M 4k+3
 
  9  
AnswerMarks Guidance
Into equation.1.1b A1F
Uses correct
reasoning to obtain
AnswerMarks Guidance
the required result.2.1 R1
Subtotal4
Question total10
QMarking instructions AO 
Question 14:
--- 14(a) ---
14(a) | Obtains 12 – 2k | 1.1b | B1 | M = 5(15 – 2k – 3) – 2(30 – 4k – 6) + (6 – 6)
= 12 – 2k
Cofactors are
–2k+12 4k –24 0 
 
–9 23 –1
 
  4k+3 –10k –9 3  
–2k+12 –9 4k+3 
1  
M –1 = 4k –24 23 –10k –9
 
12–2k
  0 −1 3  
Obtains matrix of
minors/cofactors with
at least four correct
elements.
PI by transposed
form.
Condone overall sign
error on each
element. | 1.1a | M1
Obtains matrix of
minors/cofactors with
at least seven correct
elements.
PI transposed form.
Condone overall sign
error on each
element. | 1.1a | M1
Obtains correct matrix
of minors/cofactors.
PI transposed form.
Condone overall sign
error on one element. | 1.1a | M1
Obtains fully correct,
simplified answer. | 1.1b | A1
Subtotal | 5
Q | Marking Instructions | AO | Marks | Typical Solution
--- 14(b) ---
14(b) | Obtains k ≠ 6
Follow through their
determinant. | 1.1b | B1F | k ≠ 6
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c) ---
14(c) | Uses their M –1 to form
a product to find the
solution set.
Must include
 1 
 
4k+3
 
  9  
or
Obtains
 1 
M –1 4k+3 
 
  9  
for a particular value
of k | 3.1a | M1 | x –2k+12 –9 4k+3  1 
  1   
y = 4k–24 23 –10k–9 4k+3
    
12–2k
 z    0 −1 3     9  
–2k+12–36k–27+36k+27
1  
= 4k–24+92k+69–90k–81
 
12–2k
  –4k–3+27  
 –2k+12
1  
= –36+6k
 
12–2k
 24–4k
 
 1 
 
= –3
 
  2  
which is independent of k
Obtains at least one
correct component
from their M –1, can be
unsimplified.
or
Obtains correct
 1 
–1  
M 4k+3
 
  9  
for their k | 1.1b | A1F
Obtains at least two
correct components
from their M –1 (can be
unsimplified).
or
Correctly substitutes
their
 1 
–1  
M 4k+3
 
  9  
Into equation. | 1.1b | A1F
Uses correct
reasoning to obtain
the required result. | 2.1 | R1
Subtotal | 4
Question total | 10
Q | Marking instructions | AO | Marks | Typical solution
The matrix $\mathbf{M}$ is defined as
$$\mathbf{M} = \begin{bmatrix} 5 & 2 & 1 \\ 6 & 3 & 2k + 3 \\ 2 & 1 & 5 \end{bmatrix}$$

where $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item Given that $\mathbf{M}$ is a non-singular matrix, find $\mathbf{M}^{-1}$ in terms of $k$
[5 marks]

\item State any restrictions on the value of $k$
[1 mark]

\item Using your answer to part (a), show that the solution to the set of simultaneous equations below is independent of the value of $k$

$5x + 2y + z = 1$
$6x + 3y + (2k + 3)z = 4k + 3$
$2x + y + 5z = 9$
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q14 [10]}}
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