| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvectors given eigenvalue |
| Difficulty | Standard +0.8 This FP3 question requires understanding that eigenvectors satisfy Mv = λv, then solving simultaneous equations from matrix multiplication to find three unknowns (a,b,c) and two eigenvalues. Part (b) involves standard 3×3 determinant and matrix inversion. While systematic, it requires careful algebraic manipulation across multiple steps and solid understanding of eigenvector properties, placing it moderately above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1&1&a\\2&b&c\\-1&0&1\end{pmatrix}\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}1+a\\b+c\\1\end{pmatrix}=\lambda_1\begin{pmatrix}0\\1\\1\end{pmatrix}\), so \(a=-1\), \(\lambda_1=1\) | M1, A1, A1 | M1: Multiplies matrix with first eigenvector and sets equal to \(\lambda_1\) times eigenvector. A1: Deduces \(a=-1\). A1: Deduces \(\lambda_1=1\) |
| \(\begin{pmatrix}1&1&a\\2&b&c\\-1&0&1\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1-a\\2-c\\-2\end{pmatrix}=\lambda_2\begin{pmatrix}1\\0\\-1\end{pmatrix}\), so \(c=2\), \(\lambda_2=2\) | M1, A1, A1 | M1: Multiplies matrix with second eigenvector and sets equal to \(\lambda_2\) times eigenvector. A1: Deduces \(c=2\). A1: Deduces \(\lambda_2=2\) |
| \(b+c=\lambda_1\) so \(b=-1\) | M1A1 | M1: Uses \(b+c=\lambda_1\) with their \(\lambda_1\) to find \(b\). Must have equation in \(b\) and \(c\) from first eigenvector. A1: \(b=-1\) |
| \((a=-1, b=-1, c=2, \lambda_1=1, \lambda_2=2)\) | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\det P = -d-1\) | B1 | Allow \(1-d-2\) or \(1-(2+d)\). A correct (possibly unsimplified) determinant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{P}^T=\begin{pmatrix}1&2&-1\\1&1&0\\0&d&1\end{pmatrix}\) or minors \(\begin{pmatrix}1&d+2&1\\1&1&1\\d&d&-1\end{pmatrix}\) | B1 | A correct first step |
| cofactors \(\begin{pmatrix}1&-2-d&1\\-1&1&-1\\d&-d&-1\end{pmatrix}\) | ||
| \(\frac{1}{-d-1}\begin{pmatrix}1&-1&d\\-2-d&1&-d\\1&-1&-1\end{pmatrix}\) | M1 A1 A1 | M1: Full attempt at inverse including reciprocal of determinant; at least 6 correct elements. A1: Two rows or two columns correct (ignoring determinant). M0A1A0 or M0A1A1 not possible. A1: Fully correct inverse |
| (5) | ||
| Total 13 |
# Question 5:
## Part (a)(i) & (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&1&a\\2&b&c\\-1&0&1\end{pmatrix}\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}1+a\\b+c\\1\end{pmatrix}=\lambda_1\begin{pmatrix}0\\1\\1\end{pmatrix}$, so $a=-1$, $\lambda_1=1$ | M1, A1, A1 | M1: Multiplies matrix with first eigenvector and sets equal to $\lambda_1$ times eigenvector. A1: Deduces $a=-1$. A1: Deduces $\lambda_1=1$ |
| $\begin{pmatrix}1&1&a\\2&b&c\\-1&0&1\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1-a\\2-c\\-2\end{pmatrix}=\lambda_2\begin{pmatrix}1\\0\\-1\end{pmatrix}$, so $c=2$, $\lambda_2=2$ | M1, A1, A1 | M1: Multiplies matrix with second eigenvector and sets equal to $\lambda_2$ times eigenvector. A1: Deduces $c=2$. A1: Deduces $\lambda_2=2$ |
| $b+c=\lambda_1$ so $b=-1$ | M1A1 | M1: Uses $b+c=\lambda_1$ with their $\lambda_1$ to find $b$. Must have equation in $b$ and $c$ from first eigenvector. A1: $b=-1$ |
| $(a=-1, b=-1, c=2, \lambda_1=1, \lambda_2=2)$ | | (8) |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\det P = -d-1$ | B1 | Allow $1-d-2$ or $1-(2+d)$. A correct (possibly unsimplified) determinant |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P}^T=\begin{pmatrix}1&2&-1\\1&1&0\\0&d&1\end{pmatrix}$ or minors $\begin{pmatrix}1&d+2&1\\1&1&1\\d&d&-1\end{pmatrix}$ | B1 | A correct first step |
| cofactors $\begin{pmatrix}1&-2-d&1\\-1&1&-1\\d&-d&-1\end{pmatrix}$ | | |
| $\frac{1}{-d-1}\begin{pmatrix}1&-1&d\\-2-d&1&-d\\1&-1&-1\end{pmatrix}$ | M1 A1 A1 | M1: Full attempt at inverse including reciprocal of determinant; at least 6 correct elements. A1: Two rows or two columns correct (ignoring determinant). **M0A1A0 or M0A1A1 not possible.** A1: Fully correct inverse |
| | | (5) |
| | | **Total 13** |
---\begin{enumerate}
\item The matrix $\mathbf { M }$ is given by
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & 1 & a \\
2 & b & c \\
- 1 & 0 & 1
\end{array} \right) , \text { where } a , b \text { and } c \text { are constants. }$$
(a) Given that $\mathbf { j } + \mathbf { k }$ and $\mathbf { i } - \mathbf { k }$ are two of the eigenvectors of $\mathbf { M }$, find\\
(i) the values of $a , b$ and $c$,\\
(ii) the eigenvalues which correspond to the two given eigenvectors.\\
(b) The matrix $\mathbf { P }$ is given by
$$\mathbf { P } = \left( \begin{array} { r r r }
1 & 1 & 0 \\
2 & 1 & d \\
- 1 & 0 & 1
\end{array} \right) \text {, where } d \text { is constant, } d \neq - 1$$
Find\\
(i) the determinant of $\mathbf { P }$ in terms of $d$,\\
(ii) the matrix $\mathbf { P } ^ { - 1 }$ in terms of $d$.\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q5 [13]}}