AQA Further Paper 1 2023 June — Question 10 12 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind inverse then solve system
DifficultyChallenging +1.2 This is a multi-part Further Maths matrix question requiring understanding of image/range, determinants, and matrix inverses. Part (a) requires recognizing that all columns must satisfy the plane equation (moderate insight), parts (b)(i-ii) involve standard determinant calculation and cofactor matrix methods, and (b)(iii) applies the inverse to solve a system. While it spans multiple techniques and requires some conceptual understanding of linear transformations, each individual step follows well-established procedures taught in Further Maths, making it moderately above average difficulty but not requiring exceptional insight.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix
The matrix M is defined as $$\mathbf{M} = \begin{pmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ 1 & 2 & c \end{pmatrix}$$ where \(c\) is a real number.
  1. The linear transformation T is represented by the matrix \(\mathbf{M}\) Show that, for one particular value of \(c\), the image under T of every point lies in the plane $$x + 5y + 3z = 0$$ State the value of \(c\) for which this occurs. [3 marks]
  2. It is given that M is a non-singular matrix.
    1. State any restrictions on the value of \(c\) [2 marks]
    2. Find \(\mathbf{M}^{-1}\) in terms of \(c\) [4 marks]
    3. Using your answer from part (b)(ii), solve $$2x - y + z = -3$$ $$-x - y - 2z = -6$$ $$x + 2y + 4z = 13$$ [3 marks]
Question 10:
AnswerMarks
10(a)Correctly obtains the three
components of the image of a
general point under M in terms
AnswerMarks Guidance
of cor using c =31.1b B1
    
y′ = −1 −1 −2 y
    
z′ 1 2 cz
2x−y+z
 
= −x−y−2z
 
x+2y+cz
x'+5y'+3z'=(2x−y+z)+5(−x−y−2z)+3(x+2y+cz)
=(2−5+3)x+(−1−5+6)y+(1−10+3c)z
=(−9+3c)z
So every image point lies in the
plane x+5y+3z =0 when c=3
Substitutes their components
AnswerMarks Guidance
into x+5y+3z1.1a M1
Completes a reasoned
argument to show that every
image point lies in the plane
AnswerMarks Guidance
x+5y+3z =0 when c=32.1 R1
Subtotal3
QMarking instructions AO
AnswerMarks Guidance
10(b)(i)Obtains an expression for M
Can be seen in (a)1.1a M1
M
=−3c+9
∴c≠3
AnswerMarks Guidance
Deduces correct restriction on c2.2a A1
Subtotal2
QMarking instructions AO
AnswerMarks
10(b)(ii)Obtains matrix of
minors/cofactors with at least
four correct elements
PI by transposed form.
Condone overall sign error on
AnswerMarks Guidance
each element.1.1a M1
 
minors −c−2 2c−1 5
 
  3 −3 −3 
−c+4 c−2 −1
 
Cofactors c+2 2c−1 −5
 
  3 3 −3 
−c+4 c+2 3 
1  
M−1 = c−2 2c−1 3
−3c+9  
  −1 −5 −3 
Obtains matrix of
minors/cofactors with at least
seven correct elements
PI by transposed form.
Condone overall sign error on
AnswerMarks Guidance
each element.1.1a M1
Obtains correct matrix of
minors/cofactors
PI by transposed form.
Condone overall sign error on
AnswerMarks Guidance
each element.1.1b A1
Deduces fully correct, simplified
AnswerMarks Guidance
answer2.2a A1
Subtotal4
QMarking instructions AO
AnswerMarks Guidance
10(b)(iii)Obtains their correct M−1using
c=4. Need not be simplified.3.1a B1F
  1   
y = 2 7 3 −6
    
−3
 z  −1 −5 −3  13
   
 3 
1  
= −9
 
−3
 −6
 
x=−1, y =3, z =2
−3
 
Forms their product M−1 −6
 
 13
AnswerMarks Guidance
Condone M−1 in terms of c1.1a M1
Completes a reasoned
argument to obtain the correct
solution.
−1
 
Do not accept r = 3
 
  2  
x −1
   
But do accept y = 3
   
 z    2  
NMS or any method NOT using
AnswerMarks Guidance
M−1vscores 0 marks2.1 R1
Subtotal3
Question total12
QMarking instructions AO 
Question 10:
--- 10(a) ---
10(a) | Correctly obtains the three
components of the image of a
general point under M in terms
of cor using c =3 | 1.1b | B1 | x′ 2 −1 1x
    
y′ = −1 −1 −2 y
    
z′ 1 2 cz
2x−y+z
 
= −x−y−2z
 
x+2y+cz
x'+5y'+3z'=(2x−y+z)+5(−x−y−2z)+3(x+2y+cz)
=(2−5+3)x+(−1−5+6)y+(1−10+3c)z
=(−9+3c)z
So every image point lies in the
plane x+5y+3z =0 when c=3
Substitutes their components
into x+5y+3z | 1.1a | M1
Completes a reasoned
argument to show that every
image point lies in the plane
x+5y+3z =0 when c=3 | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b)(i) ---
10(b)(i) | Obtains an expression for M
Can be seen in (a) | 1.1a | M1 | =2(−c+4)+1(−c+2)+1(−2+1)
M
=−3c+9
∴c≠3
Deduces correct restriction on c | 2.2a | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b)(ii) ---
10(b)(ii) | 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 | −c+4 −c+2 −1
 
minors −c−2 2c−1 5
 
  3 −3 −3 
−c+4 c−2 −1
 
Cofactors c+2 2c−1 −5
 
  3 3 −3 
−c+4 c+2 3 
1  
M−1 = c−2 2c−1 3
−3c+9  
  −1 −5 −3 
Obtains matrix of
minors/cofactors with at least
seven correct elements
PI by transposed form.
Condone overall sign error on
each element. | 1.1a | M1
Obtains correct matrix of
minors/cofactors
PI by transposed form.
Condone overall sign error on
each element. | 1.1b | A1
Deduces fully correct, simplified
answer | 2.2a | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b)(iii) ---
10(b)(iii) | Obtains their correct M−1using
c=4. Need not be simplified. | 3.1a | B1F | x  0 6 3 −3
  1   
y = 2 7 3 −6
    
−3
 z  −1 −5 −3  13
   
 3 
1  
= −9
 
−3
 −6
 
x=−1, y =3, z =2
−3
 
Forms their product M−1 −6
 
 13

Condone M−1 in terms of c | 1.1a | M1
Completes a reasoned
argument to obtain the correct
solution.
−1
 
Do not accept r = 3
 
  2  
x −1
   
But do accept y = 3
   
 z    2  
NMS or any method NOT using
M−1vscores 0 marks | 2.1 | R1
Subtotal | 3
Question total | 12
Q | Marking instructions | AO | Marks | Typical solution
The matrix M is defined as
$$\mathbf{M} = \begin{pmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ 1 & 2 & c \end{pmatrix}$$

where $c$ is a real number.

\begin{enumerate}[label=(\alph*)]
\item The linear transformation T is represented by the matrix $\mathbf{M}$

Show that, for one particular value of $c$, the image under T of every point lies in the plane
$$x + 5y + 3z = 0$$

State the value of $c$ for which this occurs.
[3 marks]

\item It is given that M is a non-singular matrix.

\begin{enumerate}[label=(\roman*)]
\item State any restrictions on the value of $c$
[2 marks]

\item Find $\mathbf{M}^{-1}$ in terms of $c$
[4 marks]

\item Using your answer from part (b)(ii), solve
$$2x - y + z = -3$$
$$-x - y - 2z = -6$$
$$x + 2y + 4z = 13$$
[3 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2023 Q10 [12]}}
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