AQA Further Paper 1 2023 June — Question 6 11 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind P and D for diagonalization / matrix powers
DifficultyChallenging +1.2 This is a structured Further Maths eigenvalue/eigenvector question with clear scaffolding. While it involves multiple steps (finding eigenvalues, determining unknown parameters, diagonalization), each part follows standard procedures: using Mv=λv to find unknowns, then constructing the diagonalization. The given information (two eigenvectors and one eigenvalue) makes this more routine than discovering everything from scratch. Typical for Further Maths but not requiring exceptional insight.
Spec4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
The matrix M is given by $$\mathbf{M} = \frac{1}{10} \begin{pmatrix} a & a & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{pmatrix}$$ where \(a\) is a real number. The vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are eigenvectors of \(\mathbf{M}\) The corresponding eigenvalues are \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\) respectively. It is given that \(\lambda_2 = 1\) and \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) and \(\mathbf{v}_3 = \begin{pmatrix} c \\ 0 \\ 1 \end{pmatrix}\), where \(c\) is an integer.
    1. Find the value of \(\lambda_1\) [2 marks]
    2. Find the value of \(a\) [2 marks]
  2. Find the integer \(c\) and the value of \(\lambda_3\) [4 marks]
  3. Find matrices \(\mathbf{U}\), \(\mathbf{D}\) and \(\mathbf{U}^{-1}\), such that \(\mathbf{D}\) is diagonal and \(\mathbf{M} = \mathbf{UDU}^{-1}\) [3 marks]
Question 6:
AnswerMarks
6(a)(i)Uses an appropriate
method to obtain the value
of λ
AnswerMarks Guidance
11.1a M1
1    1    
0 10 0 0 = 0 =λ 0
     1 
10 10
 9 14 −13   3    −30    3 
λ=−1
1
Obtains the correct value of
λ
AnswerMarks Guidance
11.1b A1
Subtotal2
QMarking instructions AO
AnswerMarks
6(a)(ii)Uses eigenvector definition
to set up a linear equation
AnswerMarks Guidance
in a1.1a M1
a =8
Deduces the correct value
AnswerMarks Guidance
of a2.2a A1
Subtotal2
QMarking instructions AO
AnswerMarks
6(b)Uses an appropriate
method to find λorc
PI by λ =0.5
AnswerMarks Guidance
33.1a M1
1    1    
0 10 0 0 = 0 =λ 0
     3 
10 10
 9 14 −13   1   9c−13   1 
1 (8c−6)=λc and 1 (9c−13)=λ
10 3 10 3
1 (8c−6)= c (9c−13)
10 10
9c2−21c+6=0
c=2 and c=1(reject as c∈)
3
c=2 and λ =0.5
3
Forms an equation in λorc
PI by λ =0.5
AnswerMarks Guidance
31.1a M1
Finds the correct λ
AnswerMarks Guidance
31.1b A1
Deduces the correct c
NMS = 0/4 if λwrong
AnswerMarks Guidance
32.2a A1
Subtotal4
QMarking instructions AO
AnswerMarks Guidance
6(c)Deduces a correct U, ft
their c only. Condone “c”2.2a B1F
 
U= 0 1 0
 
 3 1 1 
−1 0 0 
 
D= 0 1 0
 
  0 0 0.5 
−1 −1 2 
1 
U-1 = 0 5 0
 
5
  3 −2 −1 
Deduces the value of D, ft
their λ and λ .
1 3
Must be compatible with
their U. Condone"λ"and
1
"λ"
AnswerMarks Guidance
32.2a B1F
-1
AnswerMarks Guidance
Obtains U , ft their U1.1b B1F
Subtotal3
Question total11
QMarking instructions AO 
Question 6:
--- 6(a)(i) ---
6(a)(i) | Uses an appropriate
method to obtain the value
of λ
1 | 1.1a | M1 | a a −61 a−18 1
1    1    
0 10 0 0 = 0 =λ 0
     1 
10 10
 9 14 −13   3    −30    3 
λ=−1
1
Obtains the correct value of
λ
1 | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(a)(ii) ---
6(a)(ii) | Uses eigenvector definition
to set up a linear equation
in a | 1.1a | M1 | 10=18−a
a =8
Deduces the correct value
of a | 2.2a | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Uses an appropriate
method to find λorc
PI by λ =0.5
3 | 3.1a | M1 | 8 8 −6c 8c−6 c
1    1    
0 10 0 0 = 0 =λ 0
     3 
10 10
 9 14 −13   1   9c−13   1 
1 (8c−6)=λc and 1 (9c−13)=λ
10 3 10 3
1 (8c−6)= c (9c−13)
10 10
9c2−21c+6=0
c=2 and c=1(reject as c∈)
3
c=2 and λ =0.5
3
Forms an equation in λorc
PI by λ =0.5
3 | 1.1a | M1
Finds the correct λ
3 | 1.1b | A1
Deduces the correct c
NMS = 0/4 if λwrong
3 | 2.2a | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 6(c) ---
6(c) | Deduces a correct U, ft
their c only. Condone “c” | 2.2a | B1F | 1 1 2
 
U= 0 1 0
 
 3 1 1 
−1 0 0 
 
D= 0 1 0
 
  0 0 0.5 
−1 −1 2 
1 
U-1 = 0 5 0
 
5
  3 −2 −1 
Deduces the value of D, ft
their λ and λ .
1 3
Must be compatible with
their U. Condone"λ"and
1
"λ"
3 | 2.2a | B1F
-1
Obtains U , ft their U | 1.1b | B1F
Subtotal | 3
Question total | 11
Q | Marking instructions | AO | Marks | Typical solution
The matrix M is given by
$$\mathbf{M} = \frac{1}{10} \begin{pmatrix} a & a & -6 \\ 0 & 10 & 0 \\ 9 & 14 & -13 \end{pmatrix}$$

where $a$ is a real number.

The vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ are eigenvectors of $\mathbf{M}$

The corresponding eigenvalues are $\lambda_1$, $\lambda_2$, and $\lambda_3$ respectively.

It is given that $\lambda_2 = 1$ and $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\mathbf{v}_3 = \begin{pmatrix} c \\ 0 \\ 1 \end{pmatrix}$,

where $c$ is an integer.

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\lambda_1$
[2 marks]

\item Find the value of $a$
[2 marks]
\end{enumerate}

\item Find the integer $c$ and the value of $\lambda_3$
[4 marks]

\item Find matrices $\mathbf{U}$, $\mathbf{D}$ and $\mathbf{U}^{-1}$, such that $\mathbf{D}$ is diagonal and $\mathbf{M} = \mathbf{UDU}^{-1}$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2023 Q6 [11]}}
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