| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 18 |
| 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 Further Maths eigenvalue question covering routine techniques: finding eigenvalues/eigenvectors of a 2×2 matrix, writing the diagonalization, verifying an eigenvalue by substitution, factoring a cubic, and using the Cayley-Hamilton theorem. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow\) eigenvector is \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) o.e. | M1, A1, M1, A1, A1 | Forming characteristic polynomial |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}\) | B1ñ, B1ñ | Do not fit \(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\) as eigenvector |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda^2 + \lambda + 2 = 0 \Rightarrow (\lambda + \frac{1}{2})^2 + \frac{7}{4} = 0 \Rightarrow\) no real roots | B1, M1, A1, A1(ag) | Or showing that (λ − 5) is a factor |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x = -4, y = 2, z = 8\) | B1, B1, B2 | Allow 5\(\begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}\) isw |
| Answer | Marks | Guidance |
|---|---|---|
| and B⁴ = 4B³ + 3B² + 10B = 4(4B² + 3B + 10I) + 3B² + 10B = 19B² + 22B + 40I | M1, M1, A1(ag) | Idea of λ ↔ B. Condone omitted I |
### 3(a)(i)
Characteristic equation is $(6-\lambda)(-1-\lambda) + 12 = 0$
$\Rightarrow \lambda^2 - 5\lambda + 6 = 0$
$\Rightarrow \lambda = 2, 3$
When $\lambda = 2, \begin{pmatrix} 4 & -3 \\ 4 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
$\Rightarrow 4x - 3y = 0$
$\Rightarrow$ eigenvector is $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ o.e.
When $\lambda = 3, \begin{pmatrix} 3 & -3 \\ 4 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
$\Rightarrow x - y = 0$
$\Rightarrow$ eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ o.e. | M1, A1, M1, A1, A1 | Forming characteristic polynomial | | (A − λI)x = (λ)x M0 below | At least one equation relating x and y | For either λ = 2 or λ = 3
### 3(a)(ii)
$\mathbf{P} = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}$
$\mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ | B1ñ, B1ñ | Do not fit $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ as eigenvector | Columns must correspond | Both fits must be of numerical values; If one matrix diagonal, condone matrices not identified as **P** and **D**
### 3(b)(i)
$\lambda^3 - 4\lambda^2 - 3\lambda - 10 = 0 \Rightarrow \lambda = 5$ eigenvalue
$\lambda^3 - 4\lambda^2 - 3\lambda - 10 = (\lambda - 5)(\lambda^2 + \lambda + 2)$
$\lambda^2 + \lambda + 2 = 0 \Rightarrow (\lambda + \frac{1}{2})^2 + \frac{7}{4} = 0 \Rightarrow$ no real roots | B1, M1, A1, A1(ag) | Or showing that (λ − 5) is a factor | Obtaining quadratic factor | Correct quadratic factor | Correctly showing a correct quadratic equation has no real roots e.g. $b^2 - 4ac = 1 - 8$ or correct use of quadratic formula
### 3(b)(ii)
$\mathbf{B}\begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} = 5\begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -10 \\ 5 \\ 20 \end{pmatrix}$
$\mathbf{B}^2\begin{pmatrix} -2 \\ -2 \\ -8 \end{pmatrix} = 5^2\begin{pmatrix} -2 \\ -2 \\ -8 \end{pmatrix} = \begin{pmatrix} -100 \\ -50 \\ -200 \end{pmatrix}$
$\mathbf{B}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -20 \\ 10 \\ 40 \end{pmatrix} \Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ 8 \end{pmatrix}$
$\Rightarrow x = -4, y = 2, z = 8$ | B1, B1, B2 | Allow 5$\begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$ isw | Allow 25$\begin{pmatrix} -2 \\ -2 \\ -8 \end{pmatrix}$ or 5²$\begin{pmatrix} -2 \\ -2 \\ -8 \end{pmatrix}$ o.e. | Accept vector form | Give B1 for two correct unknowns
### 3(b)(iii)
C-H ⇒ **B**³ − 4**B**² − 3**B** − 10I = 0
$\Rightarrow \mathbf{B}^3 = 4\mathbf{B}^2 + 3\mathbf{B} + 10\mathbf{I}$
and **B**⁴ = 4**B**³ + 3**B**² + 10**B** = 4(4**B**² + 3**B** + 10I) + 3**B**² + 10**B** = 19**B**² + 22**B** + 40I | M1, M1, A1(ag) | Idea of λ ↔ **B**. Condone omitted **I** | Multiplying by **B** and substituting for **B**³ | Completion www | Condone use of M throughout
---3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the eigenvalues and corresponding eigenvectors for the matrix $\mathbf { A }$, where
$$\mathbf { A } = \left( \begin{array} { l l }
6 & - 3 \\
4 & - 1
\end{array} \right)$$
\item Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } = \mathbf { P D P } ^ { - 1 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item The $3 \times 3$ matrix $\mathbf { B }$ has characteristic equation
$$\lambda ^ { 3 } - 4 \lambda ^ { 2 } - 3 \lambda - 10 = 0$$
Show that 5 is an eigenvalue of $\mathbf { B }$. Show that $\mathbf { B }$ has no other real eigenvalues.
\item An eigenvector corresponding to the eigenvalue 5 is $\left( \begin{array} { r } - 2 \\ 1 \\ 4 \end{array} \right)$.
Evaluate $\mathbf { B } \left( \begin{array} { r } - 2 \\ 1 \\ 4 \end{array} \right)$ and $\mathbf { B } ^ { 2 } \left( \begin{array} { r } 4 \\ - 2 \\ - 8 \end{array} \right)$.\\
Solve the equation $\mathbf { B } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } - 20 \\ 10 \\ 40 \end{array} \right)$ for $x , y , z$.
\item Show that $\mathbf { B } ^ { 4 } = 19 \mathbf { B } ^ { 2 } + 22 \mathbf { B } + 40 \mathbf { I }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2014 Q3 [18]}}