| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues of 3×3 matrix |
| Difficulty | Standard +0.3 This is a standard FP2 matrices question covering characteristic equations, eigenvalues/eigenvectors, Cayley-Hamilton theorem, and matrix inversion. All parts follow routine procedures: (i) is direct computation of det(λI-M), (ii) applies standard eigenvalue methods with a hint about solving equations, (iii) states a theorem and performs algebraic manipulation, (iv) substitutes to find M^(-1). While it requires multiple techniques and careful calculation, there are no novel insights required—it's a textbook-style multi-part question slightly easier than average due to the structured guidance. |
| Spec | 4.03j Determinant 3x3: calculation4.03k Determinant 3x3: volume scale factor4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(M - \lambda I = \begin{pmatrix} 3-\lambda & 1 & -2 \\ 0 & -1-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{pmatrix}\) | M1 | Attempt at det(M - λI) with all elements present. Allow sign errors. Unsimplified. Allow signs reversed. Condone omission of = 0 |
| \(\det(M - \lambda I) = (3-\lambda)(-1-\lambda)(1-\lambda) + 2[2(-1-\lambda)]\) | A1 | |
| \(= (3-\lambda)(\lambda^2 - 1) + 4(-1-\lambda)\) | B1 | |
| \(\Rightarrow \lambda^3 - 3\lambda^2 + 3\lambda + 7 = 0\) | B1 | |
| \(\det M = -7\) | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(\lambda) = \lambda^3 - 3\lambda^2 + 3\lambda + 7\) | B1 | Showing −1 satisfies a correct characteristic equation |
| \(f(-1) = -1 - 3 - 3 + 7 = 0 \Rightarrow -1\) eigenvalue | M1 | Obtaining quadratic factor |
| \(f(\lambda) = (\lambda + 1)(\lambda^2 - 4\lambda + 7)\) | A1 | |
| \(\lambda^2 - 4\lambda + 7 = (\lambda - 2)^2 + 3 \geq 3\) so no real roots | A1 | |
| \((M - \lambda I)s = 0, \lambda = -1\) | ||
| \(\begin{pmatrix} 4 & 1 & -2 \\ 0 & 0 & 0 \\ 2 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) | M1 | Obtaining equations relating x, y and z. Obtaining equations relating two variables to a third. Dep. on first M1 |
| \(\Rightarrow 4x + y - 2z = 0\) | M1 | Obtaining equations relating x, y and z |
| \(2x + 2z = 0\) | ||
| \(\Rightarrow x = -z\) | M1 | |
| \(y = 2z - 4x = 2z + 4z = 6z\) | ||
| \(\Rightarrow s = \begin{pmatrix} -1 \\ 6 \\ 1 \end{pmatrix}\) | A1 | Or any non-zero multiple |
| \(\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -0.1 \\ 0.6 \\ 0.1 \end{pmatrix}\) | M1 | Solution by any method, e.g. use of multiple of s, but M0 if it is itself quoted without further work |
| \(\Rightarrow x = 0.1, y = -0.6, z = -0.1\) | A2 | Give A1 if any two correct |
| Answer | Marks | Guidance |
|---|---|---|
| C.H.: A matrix satisfies its own characteristic equation | B1 | Idea of \(\lambda \mapsto M\) |
| \(\Rightarrow M^3 - 3M^2 + 3M + 7I = 0\) | B1 (ag) | Must be derived www. Condone omitted I. Multiplying by \(M^{-1}\) |
| \(\Rightarrow M^3 = 3M^2 - 3M - 7I\) | ||
| \(\Rightarrow M^2 = 3M - 3I - 7M^{-1}\) | M1 | |
| \(\Rightarrow M^{-1} = -\frac{1}{7}M^2 + \frac{3}{7}M - I\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(M^2 = \begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 2 & -8 \\ 0 & 1 & 0 \\ 8 & 2 & -3 \end{pmatrix}\) | M1 | Correct attempt to find \(M^2\) |
| \(= \frac{1}{7}\begin{pmatrix} 5 & 2 & -8 \\ 0 & 1 & 0 \\ 8 & 2 & -3 \end{pmatrix} + \frac{3}{7}\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} - \frac{1}{7}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) | M1 | Using their (iii) |
| \(= \begin{pmatrix} \frac{5}{7} & \frac{3}{7} & \frac{1}{7} \\ 0 & -1 & 0 \end{pmatrix} \text{ or } \frac{1}{7}\begin{pmatrix} 1 & 1 & 2 \\ 0 & -7 & 0 \\ -2 & -2 & 3 \end{pmatrix}\) | A1 | SC1 for answer without working |
## Part (i)
$M - \lambda I = \begin{pmatrix} 3-\lambda & 1 & -2 \\ 0 & -1-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{pmatrix}$ | M1 | Attempt at det(M - λI) with all elements present. Allow sign errors. Unsimplified. Allow signs reversed. Condone omission of = 0
$\det(M - \lambda I) = (3-\lambda)(-1-\lambda)(1-\lambda) + 2[2(-1-\lambda)]$ | A1 |
$= (3-\lambda)(\lambda^2 - 1) + 4(-1-\lambda)$ | B1 |
$\Rightarrow \lambda^3 - 3\lambda^2 + 3\lambda + 7 = 0$ | B1 |
$\det M = -7$ | | 3 marks total
## Part (ii)
$f(\lambda) = \lambda^3 - 3\lambda^2 + 3\lambda + 7$ | B1 | Showing −1 satisfies a correct characteristic equation
$f(-1) = -1 - 3 - 3 + 7 = 0 \Rightarrow -1$ eigenvalue | M1 | Obtaining quadratic factor
$f(\lambda) = (\lambda + 1)(\lambda^2 - 4\lambda + 7)$ | A1 |
$\lambda^2 - 4\lambda + 7 = (\lambda - 2)^2 + 3 \geq 3$ so no real roots | A1 |
$(M - \lambda I)s = 0, \lambda = -1$ | |
$\begin{pmatrix} 4 & 1 & -2 \\ 0 & 0 & 0 \\ 2 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ | M1 | Obtaining equations relating x, y and z. Obtaining equations relating two variables to a third. Dep. on first M1
$\Rightarrow 4x + y - 2z = 0$ | M1 | Obtaining equations relating x, y and z
$2x + 2z = 0$ | |
$\Rightarrow x = -z$ | M1 |
$y = 2z - 4x = 2z + 4z = 6z$ | |
$\Rightarrow s = \begin{pmatrix} -1 \\ 6 \\ 1 \end{pmatrix}$ | A1 | Or any non-zero multiple
$\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -0.1 \\ 0.6 \\ 0.1 \end{pmatrix}$ | M1 | Solution by any method, e.g. use of multiple of s, but M0 if it is itself quoted without further work
$\Rightarrow x = 0.1, y = -0.6, z = -0.1$ | A2 | Give A1 if any two correct | | 9 marks total
## Part (iii)
**C.H.**: A matrix satisfies its own characteristic equation | B1 | Idea of $\lambda \mapsto M$
$\Rightarrow M^3 - 3M^2 + 3M + 7I = 0$ | B1 (ag) | Must be derived www. Condone omitted I. Multiplying by $M^{-1}$
$\Rightarrow M^3 = 3M^2 - 3M - 7I$ | |
$\Rightarrow M^2 = 3M - 3I - 7M^{-1}$ | M1 |
$\Rightarrow M^{-1} = -\frac{1}{7}M^2 + \frac{3}{7}M - I$ | A1 | | 4 marks total
## Part (iv)
$M^2 = \begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 2 & -8 \\ 0 & 1 & 0 \\ 8 & 2 & -3 \end{pmatrix}$ | M1 | Correct attempt to find $M^2$
$= \frac{1}{7}\begin{pmatrix} 5 & 2 & -8 \\ 0 & 1 & 0 \\ 8 & 2 & -3 \end{pmatrix} + \frac{3}{7}\begin{pmatrix} 3 & 1 & -2 \\ 0 & -1 & 0 \\ 2 & 0 & 1 \end{pmatrix} - \frac{1}{7}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ | M1 | Using their (iii)
$= \begin{pmatrix} \frac{5}{7} & \frac{3}{7} & \frac{1}{7} \\ 0 & -1 & 0 \end{pmatrix} \text{ or } \frac{1}{7}\begin{pmatrix} 1 & 1 & 2 \\ 0 & -7 & 0 \\ -2 & -2 & 3 \end{pmatrix}$ | A1 | SC1 for answer without working | | 3 marks total
---\begin{enumerate}[label=(\roman*)]
\item Obtain the characteristic equation for the matrix $\mathbf{M}$ where
$$\mathbf{M} = \begin{pmatrix}
3 & 1 & -2 \\
6 & -1 & 0 \\
2 & 0 & 1
\end{pmatrix}.$$
Hence or otherwise obtain the value of $\det(\mathbf{M})$. [3]
\item Show that $-1$ is an eigenvalue of $\mathbf{M}$, and show that the other two eigenvalues are not real.
Find an eigenvector corresponding to the eigenvalue $-1$.
Hence or otherwise write down the solution to the following system of equations. [9]
$$3x + y - 2z = -0.1$$
$$-y = 0.6$$
$$2x + z = 0.1$$
\item State the Cayley-Hamilton theorem and use it to show that
$$\mathbf{M}^3 = 3\mathbf{M}^2 - 3\mathbf{M} - 7\mathbf{I}.$$
Obtain an expression for $\mathbf{M}^{-1}$ in terms of $\mathbf{M}^2$, $\mathbf{M}$ and $\mathbf{I}$. [4]
\item Find the numerical values of the elements of $\mathbf{M}^{-1}$, showing your working. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q2 [19]}}