Edexcel F3 2015 June — Question 3 12 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeEigenvalues and eigenvectors
DifficultyStandard +0.3 This is a structured multi-part question on eigenvalues/eigenvectors that guides students through standard procedures: finding an eigenvalue from a given eigenvector (matrix multiplication), determining k from the eigenvalue equation, finding remaining eigenvalues (characteristic equation with one known root), and solving a matrix equation. While it involves Further Maths content, each part follows routine algorithms with no novel insight required, making it slightly easier than average.
Spec4.03a Matrix language: terminology and notation
  1. \(\mathbf { M } = \left( \begin{array} { r r r } 0 & 1 & 9 \\ 1 & 4 & k \\ 1 & 0 & - 3 \end{array} \right)\), where \(k\) is a constant.
Given that \(\left( \begin{array} { r } 7 \\ 19 \\ 1 \end{array} \right)\) is an eigenvector of the matrix \(\mathbf { M }\),
  1. find the eigenvalue of \(\mathbf { M }\) corresponding to \(\left( \begin{array} { r } 7 \\ 19 \\ 1 \end{array} \right)\),
  2. show that \(k = - 7\)
  3. find the other two eigenvalues of the matrix \(\mathbf { M }\). The image of the vector \(\left( \begin{array} { c } p \\ q \\ r \end{array} \right)\) under the transformation represented by \(\mathbf { M }\) is \(\left( \begin{array} { r } - 6 \\ 21 \\ 5 \end{array} \right)\).
  4. Find the values of the constants \(p , q\) and \(r\).
Question 3:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{pmatrix}0&1&9\\1&4&k\\1&0&-3\end{pmatrix}\begin{pmatrix}7\\19\\1\end{pmatrix} = \lambda\begin{pmatrix}7\\19\\1\end{pmatrix}\)M1 Correct statement
\(7 - 3 = \lambda\) or \(28 = 7\lambda \Rightarrow \lambda = 4\)A1 Correct eigenvalue
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(7 + 4\times19 + k = 4\times19 \Rightarrow k = -7\)M1A1* M1: Uses \(y\) component to establish equation for \(k\). A1*: Correct \(k\)
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{vmatrix}-\lambda & 1 & 9\\1 & 4-\lambda & -7\\1 & 0 & -3-\lambda\end{vmatrix}=0\)
\(\lambda(4-\lambda)(3+\lambda)+(3+\lambda)-7+9(\lambda-4)=0\) or \(-7+9(\lambda-4)-(3+\lambda)[\lambda(\lambda-4)-1]\)M1A1 M1: Correct characteristic equation method (allow sign errors only). A1: Correct equation in any form
\((4-\lambda)[\lambda(3+\lambda)-1-9]=0\) NB \(\lambda^3 - \lambda^2 - 22\lambda + 40 = 0\)
\((\lambda-2)(\lambda+5)=0 \Rightarrow \lambda = 2, -5\)A1, A1 A1: \(\lambda=2\) or \(\lambda=-5\); A1: \(\lambda=2\) and \(\lambda=-5\)
Part (d) — Way 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{pmatrix}0&1&9\\1&4&-7\\1&0&-3\end{pmatrix}\begin{pmatrix}p\\q\\r\end{pmatrix}=\begin{pmatrix}q+9r\\p+4q-7r\\p-3r\end{pmatrix}\)M1 Multiplies by M to obtain a vector in terms of \(p\), \(q\) and \(r\)
\(\begin{pmatrix}q+9r\\p+4q-7r\\p-3r\end{pmatrix}=\begin{pmatrix}-6\\21\\5\end{pmatrix}\)A1 Correct equations
\(p=2,\ q=3,\ r=-1\)dM1A1 M1: Solves simultaneously to obtain at least one of \(p\), \(q\) or \(r\). Dependent on previous method mark. A1: Correct answers
Part (d) — Way 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{pmatrix}p\\q\\r\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}-6\\21\\5\end{pmatrix}=\frac{1}{40}\begin{pmatrix}12&-3&43\\4&9&-9\\4&-1&1\end{pmatrix}\begin{pmatrix}-6\\21\\5\end{pmatrix}\)M1A1 M1: Appreciation that \(p\mathbf{i}+q\mathbf{j}+r\mathbf{k}=\mathbf{M}^{-1}(-6\mathbf{i}+21\mathbf{j}+5\mathbf{k})\). A1: Correct inverse
\(\begin{pmatrix}p\\q\\r\end{pmatrix}=\frac{1}{40}\begin{pmatrix}12&-3&43\\4&9&-9\\4&-1&1\end{pmatrix}\begin{pmatrix}-6\\21\\5\end{pmatrix}=\begin{pmatrix}2\\3\\-1\end{pmatrix}\)dM1A1 M1: Multiplies inverse by given vector. Dependent on previous method mark. A1: Correct vector
Total: 12 marks
# Question 3:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}0&1&9\\1&4&k\\1&0&-3\end{pmatrix}\begin{pmatrix}7\\19\\1\end{pmatrix} = \lambda\begin{pmatrix}7\\19\\1\end{pmatrix}$ | M1 | Correct statement |
| $7 - 3 = \lambda$ or $28 = 7\lambda \Rightarrow \lambda = 4$ | A1 | Correct eigenvalue |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $7 + 4\times19 + k = 4\times19 \Rightarrow k = -7$ | M1A1* | M1: Uses $y$ component to establish equation for $k$. A1*: Correct $k$ |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}-\lambda & 1 & 9\\1 & 4-\lambda & -7\\1 & 0 & -3-\lambda\end{vmatrix}=0$ | | |
| $\lambda(4-\lambda)(3+\lambda)+(3+\lambda)-7+9(\lambda-4)=0$ or $-7+9(\lambda-4)-(3+\lambda)[\lambda(\lambda-4)-1]$ | M1A1 | M1: Correct characteristic equation method (allow sign errors only). A1: Correct equation in any form |
| $(4-\lambda)[\lambda(3+\lambda)-1-9]=0$ | | NB $\lambda^3 - \lambda^2 - 22\lambda + 40 = 0$ |
| $(\lambda-2)(\lambda+5)=0 \Rightarrow \lambda = 2, -5$ | A1, A1 | A1: $\lambda=2$ or $\lambda=-5$; A1: $\lambda=2$ and $\lambda=-5$ |

## Part (d) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}0&1&9\\1&4&-7\\1&0&-3\end{pmatrix}\begin{pmatrix}p\\q\\r\end{pmatrix}=\begin{pmatrix}q+9r\\p+4q-7r\\p-3r\end{pmatrix}$ | M1 | Multiplies by **M** to obtain a vector in terms of $p$, $q$ and $r$ |
| $\begin{pmatrix}q+9r\\p+4q-7r\\p-3r\end{pmatrix}=\begin{pmatrix}-6\\21\\5\end{pmatrix}$ | A1 | Correct equations |
| $p=2,\ q=3,\ r=-1$ | dM1A1 | M1: Solves simultaneously to obtain at least one of $p$, $q$ or $r$. **Dependent on previous method mark.** A1: Correct answers |

## Part (d) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}p\\q\\r\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}-6\\21\\5\end{pmatrix}=\frac{1}{40}\begin{pmatrix}12&-3&43\\4&9&-9\\4&-1&1\end{pmatrix}\begin{pmatrix}-6\\21\\5\end{pmatrix}$ | M1A1 | M1: Appreciation that $p\mathbf{i}+q\mathbf{j}+r\mathbf{k}=\mathbf{M}^{-1}(-6\mathbf{i}+21\mathbf{j}+5\mathbf{k})$. A1: Correct inverse |
| $\begin{pmatrix}p\\q\\r\end{pmatrix}=\frac{1}{40}\begin{pmatrix}12&-3&43\\4&9&-9\\4&-1&1\end{pmatrix}\begin{pmatrix}-6\\21\\5\end{pmatrix}=\begin{pmatrix}2\\3\\-1\end{pmatrix}$ | dM1A1 | M1: Multiplies inverse by given vector. **Dependent on previous method mark.** A1: Correct vector |

**Total: 12 marks**

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\begin{enumerate}
  \item $\mathbf { M } = \left( \begin{array} { r r r } 0 & 1 & 9 \\ 1 & 4 & k \\ 1 & 0 & - 3 \end{array} \right)$, where $k$ is a constant.
\end{enumerate}

Given that $\left( \begin{array} { r } 7 \\ 19 \\ 1 \end{array} \right)$ is an eigenvector of the matrix $\mathbf { M }$,\\
(a) find the eigenvalue of $\mathbf { M }$ corresponding to $\left( \begin{array} { r } 7 \\ 19 \\ 1 \end{array} \right)$,\\
(b) show that $k = - 7$\\
(c) find the other two eigenvalues of the matrix $\mathbf { M }$.

The image of the vector $\left( \begin{array} { c } p \\ q \\ r \end{array} \right)$ under the transformation represented by $\mathbf { M }$ is $\left( \begin{array} { r } - 6 \\ 21 \\ 5 \end{array} \right)$.\\
(d) Find the values of the constants $p , q$ and $r$.\\

\hfill \mbox{\textit{Edexcel F3 2015 Q3 [12]}}
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