Question 1

Edexcel FP2 AS Specimen — Question 1 5 marks

Exam Board	Edexcel
Module	FP2 AS (Further Pure 2 AS)
Session	Specimen
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Use Cayley-Hamilton for matrix power
Difficulty	Standard +0.3 This is a straightforward application of the Cayley-Hamilton theorem with standard matrix algebra. Part (a) requires finding a 2×2 characteristic equation (routine calculation), and part (b) is a guided proof using the theorem to express A³ in terms of A and I. The 'hence' makes it clear what method to use, requiring only algebraic manipulation rather than problem-solving insight.
Spec	4.03g Invariant points and lines 4.03h Determinant 2x2: calculation

Given that

$$A = \left( \begin{array} { l l } 3 & 1 \\ 6 & 4 \end{array} \right)$$

find the characteristic equation of the matrix $\mathbf { A }$.
Hence show that $\mathbf { A } ^ { 3 } = 43 \mathbf { A } - 42 \mathbf { I }$.

Show mark scheme Show mark scheme source

1(a)

Consider $\det \begin{pmatrix} 3-\lambda & 1 \\ 6 & 4-\lambda \end{pmatrix} = (3-\lambda)(4-\lambda) - 6$

Answer	Marks
M1	1.1b

So $\lambda^2 - 7\lambda + 6 = 0$ is characteristic equation

Answer	Marks
A1	1.1b

So $A^2 = 7A - 6I$

Answer	Marks
B1ft	1.1b

(2)

1(b)

Multiplies both sides of their equation by $A$ so $A^3 = 7A^2 - 6A$

Answer	Marks
M1	3.1a

Uses $A^3 = 7(7A - 6I) - 6A$ So $A^3 = 43A - 42I$

Answer	Marks
A1*cso	1.1b

(3)

(5 marks)

Notes:

(a)

M1: Complete method to find characteristic equation

A1: Obtains a correct three term quadratic equation – may use variable other than $\lambda$

(b)

B1ft: Uses Cayley Hamilton Theorem to produce equation replacing $\lambda$ with $A$ and constant term with constant multiple of identity matrix, $I$

M1: Multiplies equation by $A$

A1*: Replaces $A^2$ by linear expression in $A$ and achieves printed answer with no errors

# Question 1

## 1(a)

Consider $\det \begin{pmatrix} 3-\lambda & 1 \\ 6 & 4-\lambda \end{pmatrix} = (3-\lambda)(4-\lambda) - 6$

M1 | 1.1b

So $\lambda^2 - 7\lambda + 6 = 0$ is characteristic equation

A1 | 1.1b

So $A^2 = 7A - 6I$

B1ft | 1.1b

(2)

## 1(b)

Multiplies both sides of their equation by $A$ so $A^3 = 7A^2 - 6A$

M1 | 3.1a

Uses $A^3 = 7(7A - 6I) - 6A$ So $A^3 = 43A - 42I$

A1*cso | 1.1b

(3)

(5 marks)

## Notes:

### (a)

M1: Complete method to find characteristic equation

A1: Obtains a correct three term quadratic equation – may use variable other than $\lambda$

### (b)

B1ft: Uses Cayley Hamilton Theorem to produce equation replacing $\lambda$ with $A$ and constant term with constant multiple of identity matrix, $I$

M1: Multiplies equation by $A$

A1*: Replaces $A^2$ by linear expression in $A$ and achieves printed answer with no errors

Show LaTeX source

\begin{enumerate}
  \item Given that
\end{enumerate}

$$A = \left( \begin{array} { l l } 
3 & 1 \\
6 & 4
\end{array} \right)$$

(a) find the characteristic equation of the matrix $\mathbf { A }$.\\
(b) Hence show that $\mathbf { A } ^ { 3 } = 43 \mathbf { A } - 42 \mathbf { I }$.

\hfill \mbox{\textit{Edexcel FP2 AS  Q1 [5]}}

This paper (5 questions)

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Q1 5 Q2 6 Q3 8 Q4 11 Q5 10

Edexcel FP2 AS Specimen — Question 1 5 marks

Question 1

1(a)

Consider \(\det \begin{pmatrix} 3-\lambda & 1 \\ 6 & 4-\lambda \end{pmatrix} = (3-\lambda)(4-\lambda) - 6\)

So \(\lambda^2 - 7\lambda + 6 = 0\) is characteristic equation

So \(A^2 = 7A - 6I\)

(2)

1(b)

Multiplies both sides of their equation by \(A\) so \(A^3 = 7A^2 - 6A\)

Uses \(A^3 = 7(7A - 6I) - 6A\) So \(A^3 = 43A - 42I\)

(3)

(5 marks)

Notes:

(a)

M1: Complete method to find characteristic equation

A1: Obtains a correct three term quadratic equation – may use variable other than \(\lambda\)

(b)

B1ft: Uses Cayley Hamilton Theorem to produce equation replacing \(\lambda\) with \(A\) and constant term with constant multiple of identity matrix, \(I\)

M1: Multiplies equation by \(A\)

A1*: Replaces \(A^2\) by linear expression in \(A\) and achieves printed answer with no errors

This paper (5 questions)