Question 4 - A-Level Maths

Edexcel FP2 AS 2018 June — Question 4 7 marks

Exam Board	Edexcel
Module	FP2 AS (Further Pure 2 AS)
Year	2018
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find P and D for A = PDP⁻¹
Difficulty	Standard +0.3 This is a standard diagonalization problem requiring finding eigenvalues (solving a quadratic characteristic equation), finding corresponding eigenvectors (solving two simple linear systems), and constructing P and D. It's slightly easier than average because the 2×2 matrix has nice integer eigenvalues and the procedure is completely routine for FP2 students with no conceptual challenges.
Spec	4.03n Inverse 2x2 matrix 4.03o Inverse 3x3 matrix

4. $$\mathbf { A } = \left( \begin{array} { r r } 1 & 1 \\ - 2 & 4 \end{array} \right)$$ Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { D } = \mathbf { P } ^ { - 1 } \mathbf { A P }$

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Question 4:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\	\mathbf{A}-\lambda\mathbf{I}\	=0 \Rightarrow \begin{vmatrix}1-\lambda & 1\\-2 & 4-\lambda\end{vmatrix}=0 \Rightarrow (1-\lambda)(4-\lambda)+2=0\)
$\lambda_1=2,\quad \lambda_2=3$	A1	Correct eigenvalues
$\begin{pmatrix}1&1\\-2&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=2\begin{pmatrix}x\\y\end{pmatrix}$ or $\begin{pmatrix}1&1\\-2&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=3\begin{pmatrix}x\\y\end{pmatrix}$	M1	Uses at least one eigenvalue to find eigenvector
$2,\begin{pmatrix}1\\1\end{pmatrix}$ or $3,\begin{pmatrix}1\\2\end{pmatrix}$	A1	One correct eigenvalue/eigenvector pair
$2,\begin{pmatrix}1\\1\end{pmatrix}$ and $3,\begin{pmatrix}1\\2\end{pmatrix}$	A1	Both pairs correct
$\mathbf{D}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$	B1ft	Correct follow through, clearly identified as $\mathbf{D}$
$\mathbf{P}=\begin{pmatrix}1&1\\1&2\end{pmatrix}$	B1ft	$\mathbf{P}$ and $\mathbf{D}$ both correct, consistent, and identified

# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|\mathbf{A}-\lambda\mathbf{I}\|=0 \Rightarrow \begin{vmatrix}1-\lambda & 1\\-2 & 4-\lambda\end{vmatrix}=0 \Rightarrow (1-\lambda)(4-\lambda)+2=0$ | M1 | Correct strategy for finding eigenvalues |
| $\lambda_1=2,\quad \lambda_2=3$ | A1 | Correct eigenvalues |
| $\begin{pmatrix}1&1\\-2&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=2\begin{pmatrix}x\\y\end{pmatrix}$ or $\begin{pmatrix}1&1\\-2&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=3\begin{pmatrix}x\\y\end{pmatrix}$ | M1 | Uses at least one eigenvalue to find eigenvector |
| $2,\begin{pmatrix}1\\1\end{pmatrix}$ or $3,\begin{pmatrix}1\\2\end{pmatrix}$ | A1 | One correct eigenvalue/eigenvector pair |
| $2,\begin{pmatrix}1\\1\end{pmatrix}$ and $3,\begin{pmatrix}1\\2\end{pmatrix}$ | A1 | Both pairs correct |
| $\mathbf{D}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ | B1ft | Correct follow through, clearly identified as $\mathbf{D}$ |
| $\mathbf{P}=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ | B1ft | $\mathbf{P}$ and $\mathbf{D}$ both correct, consistent, and identified |

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4.

$$\mathbf { A } = \left( \begin{array} { r r } 
1 & 1 \\
- 2 & 4
\end{array} \right)$$

Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { D } = \mathbf { P } ^ { - 1 } \mathbf { A P }$

\hfill \mbox{\textit{Edexcel FP2 AS 2018 Q4 [7]}}

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Q1 5 Q2 10 Q3 10 Q4 7 Q5 8