Edexcel F3 2021 January — Question 5 9 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2021
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeOrthogonal matrix diagonalization
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring eigenvalue/eigenvector computation, finding remaining eigenvalues (likely via trace/determinant), and orthogonal diagonalization. While systematic, it demands careful matrix algebra, normalization of eigenvectors, and understanding of orthogonal matrices—significantly above standard A-level but routine for F3 students who have practiced this technique.
Spec4.03k Determinant 3x3: volume scale factor4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix

5. $$\mathbf { M } = \left( \begin{array} { r r r } 6 & - 2 & - 1 \\ - 2 & 6 & - 1 \\ - 1 & - 1 & 5 \end{array} \right)$$ Given that 8 is an eigenvalue of \(\mathbf { M }\)
  1. determine an eigenvector corresponding to the eigenvalue 8
  2. Determine the other two eigenvalues of \(\mathbf { M }\).
  3. Hence find an orthogonal matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { P } ^ { T } \mathbf { M P } = \mathbf { D }\) 5.

Question 5(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{pmatrix}6&-2&-1\\-2&6&-1\\-1&-1&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=8\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow \begin{pmatrix}-2&-2&-1\\-2&-2&-1\\-1&-1&-3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{0}\)M1 Correct method for obtaining eigenvector
\(\mathbf{i}-\mathbf{j}\)A1 Any multiple of this vector
Question 5(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{M}-\lambda\mathbf{I} = \begin{vmatrix}6-\lambda&-2&-1\\-2&6-\lambda&-1\\-1&-1&5-\lambda\end{vmatrix}\) expanding correctly
\(\lambda^3 - 17\lambda^2 + 90\lambda - 144 = 0\)M1 Solves \(\mathbf{M}-\lambda\mathbf{I}=0\) to obtain 2 different distinct real eigenvalues excluding 8
\(\lambda = 3, 6, (8)\)A1 For 3 and 6
Question 5(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{D} = \begin{pmatrix}8&0&0\\0&3&0\\0&0&6\end{pmatrix}\)B1ft Correct \(\mathbf{D}\) with distinct non-zero eigenvalues in any order. Follow through their non-zero 3 and 6
For \(\lambda=3\): eigenvector \(\mathbf{v}_2 = k\begin{pmatrix}1\\1\\1\end{pmatrix}\); for \(\lambda=6\): eigenvector \(\mathbf{v}_3 = k\begin{pmatrix}1\\1\\-2\end{pmatrix}\)M1 Attempts eigenvectors for other 2 distinct eigenvalues not including 8
Forms complete \(\mathbf{P}\) from normalised eigenvectorsM1 Forms complete \(\mathbf{P}\) from normalised eigenvectors using eigenvector from (a) and other 2 eigenvectors
\(\mathbf{P} = \begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{6}}\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{6}}\\0&\frac{1}{\sqrt{3}}&-\frac{2}{\sqrt{6}}\end{pmatrix}\)A1 All fully correct, consistent and correctly labelled
# Question 5(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}6&-2&-1\\-2&6&-1\\-1&-1&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=8\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow \begin{pmatrix}-2&-2&-1\\-2&-2&-1\\-1&-1&-3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{0}$ | M1 | Correct method for obtaining eigenvector |
| $\mathbf{i}-\mathbf{j}$ | A1 | Any multiple of this vector |

# Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\mathbf{M}-\lambda\mathbf{I}| = \begin{vmatrix}6-\lambda&-2&-1\\-2&6-\lambda&-1\\-1&-1&5-\lambda\end{vmatrix}$ expanding correctly | M1 | Correct attempt at determinant of $\mathbf{M}-\lambda\mathbf{I}$; terms with single underlining correct with correct signs |
| $\lambda^3 - 17\lambda^2 + 90\lambda - 144 = 0$ | M1 | Solves $\mathbf{M}-\lambda\mathbf{I}=0$ to obtain 2 different distinct real eigenvalues excluding 8 |
| $\lambda = 3, 6, (8)$ | A1 | For 3 and 6 |

# Question 5(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{D} = \begin{pmatrix}8&0&0\\0&3&0\\0&0&6\end{pmatrix}$ | B1ft | Correct $\mathbf{D}$ with distinct non-zero eigenvalues in any order. Follow through their non-zero 3 and 6 |
| For $\lambda=3$: eigenvector $\mathbf{v}_2 = k\begin{pmatrix}1\\1\\1\end{pmatrix}$; for $\lambda=6$: eigenvector $\mathbf{v}_3 = k\begin{pmatrix}1\\1\\-2\end{pmatrix}$ | M1 | Attempts eigenvectors for other 2 distinct eigenvalues not including 8 |
| Forms complete $\mathbf{P}$ from normalised eigenvectors | M1 | Forms complete $\mathbf{P}$ from normalised eigenvectors using eigenvector from (a) and other 2 eigenvectors |
| $\mathbf{P} = \begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{6}}\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{6}}\\0&\frac{1}{\sqrt{3}}&-\frac{2}{\sqrt{6}}\end{pmatrix}$ | A1 | All fully correct, consistent and correctly labelled |
5.

$$\mathbf { M } = \left( \begin{array} { r r r } 
6 & - 2 & - 1 \\
- 2 & 6 & - 1 \\
- 1 & - 1 & 5
\end{array} \right)$$

Given that 8 is an eigenvalue of $\mathbf { M }$
\begin{enumerate}[label=(\alph*)]
\item determine an eigenvector corresponding to the eigenvalue 8
\item Determine the other two eigenvalues of $\mathbf { M }$.
\item Hence find an orthogonal matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { P } ^ { T } \mathbf { M P } = \mathbf { D }$\\
5.\\

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\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2021 Q5 [9]}}