| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Use Cayley-Hamilton for inverse |
| Difficulty | Standard +0.3 This is a standard Further Pure 2 eigenvalue question with routine steps: finding characteristic equation (given result to verify), factorizing a cubic, finding eigenvectors, and applying Cayley-Hamilton theorem. All parts follow textbook procedures with no novel insight required. The Cayley-Hamilton application is mechanical substitution. Slightly easier than average A-level due to the structured guidance and verification format. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}-\lambda\mathbf{I} = \begin{pmatrix}3-\lambda & -1 & 2\\-4 & 3-\lambda & 2\\2 & 1 & -1-\lambda\end{pmatrix}\) | ||
| \(\det(\mathbf{M}-\lambda\mathbf{I}) = (3-\lambda)[(3-\lambda)(-1-\lambda)-2]+1[-4(-1-\lambda)-4]+2[-4-2(3-\lambda)]\) | M1 | Obtaining \(\det(\mathbf{M}-\lambda\mathbf{I})\) |
| Any correct form | A1 | |
| \(=(3-\lambda)(\lambda^2-2\lambda-5)+4\lambda+2(2\lambda-10)\) | M1 | Multiplying out. Dependent on first M1 |
| \(\Rightarrow \lambda^3-5\lambda^2-7\lambda+35=0\) | E1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda^3-5\lambda^2-7\lambda+35=0\) | M1 | Factorising, obtaining a quadratic. If M0, give B1 for substituting \(\lambda=5\) |
| \(\Rightarrow (\lambda-5)(\lambda^2-7)=0\) | A1 | Correct quadratic |
| M1 | Solving quadratic | |
| \(\lambda = \pm\sqrt{7}\) | A1 | Allow 2.65 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda=5 \Rightarrow \begin{pmatrix}-2&-1&2\\-4&-2&2\\2&1&-6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | ||
| \(\Rightarrow -2x-y+2z=0;\; -4x-2y+2z=0;\; 2x+y-6z=0\) | M1 | Two independent equations. Need to multiply out, unless implied by later work |
| \(\Rightarrow z=0,\; y=-2x\) | M1 | Obtaining a non-zero eigenvector |
| \(\Rightarrow\) eigenvector is \(\begin{pmatrix}1\\-2\\0\end{pmatrix}\) | A1 | |
| \(\Rightarrow\) eigenvector of unit length is \(\frac{1}{\sqrt{5}}\begin{pmatrix}1\\-2\\0\end{pmatrix}\) | A1ft | \(\frac{1}{\sqrt{5}}\) f.t. their eigenvector |
| \(\mathbf{M}^2\mathbf{v}\) has magnitude 25 in direction of \(\mathbf{v}\) | B1 | Both magnitudes c.a.o. May be given as column vectors |
| \(\mathbf{M}^{-1}\mathbf{v}\) has magnitude \(\frac{1}{5}\) in direction of \(\mathbf{v}\) | B1 | Directions c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda^3-5\lambda^2-7\lambda+35=0 \Rightarrow \mathbf{M}^3-5\mathbf{M}^2-7\mathbf{M}+35\mathbf{I}=\mathbf{0}\) | M1 | Using Cayley-Hamilton Theorem. Condone omitted \(\mathbf{I}\) |
| \(\Rightarrow \mathbf{M}^4 = 5\mathbf{M}^3+7\mathbf{M}^2-35\mathbf{M}\) | A1 | Correct expression involving \(\mathbf{M}^4\) and non-negative powers of \(\mathbf{M}\) |
| \(= 5(5\mathbf{M}^2+7\mathbf{M}-35\mathbf{I})+7\mathbf{M}^2-35\mathbf{M}\) | M1 | Substituting for \(\mathbf{M}^3\) and obtaining expression in required form |
| \(= 32\mathbf{M}^2-175\mathbf{I}\) | A1 | \(a=32,\; b=0,\; c=-175\) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}-\lambda\mathbf{I} = \begin{pmatrix}3-\lambda & -1 & 2\\-4 & 3-\lambda & 2\\2 & 1 & -1-\lambda\end{pmatrix}$ | | |
| $\det(\mathbf{M}-\lambda\mathbf{I}) = (3-\lambda)[(3-\lambda)(-1-\lambda)-2]+1[-4(-1-\lambda)-4]+2[-4-2(3-\lambda)]$ | M1 | Obtaining $\det(\mathbf{M}-\lambda\mathbf{I})$ |
| Any correct form | A1 | |
| $=(3-\lambda)(\lambda^2-2\lambda-5)+4\lambda+2(2\lambda-10)$ | M1 | Multiplying out. Dependent on first M1 |
| $\Rightarrow \lambda^3-5\lambda^2-7\lambda+35=0$ | E1 | Answer given |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda^3-5\lambda^2-7\lambda+35=0$ | M1 | Factorising, obtaining a quadratic. If M0, give B1 for substituting $\lambda=5$ |
| $\Rightarrow (\lambda-5)(\lambda^2-7)=0$ | A1 | Correct quadratic |
| | M1 | Solving quadratic |
| $\lambda = \pm\sqrt{7}$ | A1 | Allow 2.65 or better |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda=5 \Rightarrow \begin{pmatrix}-2&-1&2\\-4&-2&2\\2&1&-6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | | |
| $\Rightarrow -2x-y+2z=0;\; -4x-2y+2z=0;\; 2x+y-6z=0$ | M1 | Two independent equations. Need to multiply out, unless implied by later work |
| $\Rightarrow z=0,\; y=-2x$ | M1 | Obtaining a non-zero eigenvector |
| $\Rightarrow$ eigenvector is $\begin{pmatrix}1\\-2\\0\end{pmatrix}$ | A1 | |
| $\Rightarrow$ eigenvector of unit length is $\frac{1}{\sqrt{5}}\begin{pmatrix}1\\-2\\0\end{pmatrix}$ | A1ft | $\frac{1}{\sqrt{5}}$ f.t. their eigenvector |
| $\mathbf{M}^2\mathbf{v}$ has magnitude 25 in direction of $\mathbf{v}$ | B1 | Both magnitudes c.a.o. May be given as column vectors |
| $\mathbf{M}^{-1}\mathbf{v}$ has magnitude $\frac{1}{5}$ in direction of $\mathbf{v}$ | B1 | Directions c.a.o. |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda^3-5\lambda^2-7\lambda+35=0 \Rightarrow \mathbf{M}^3-5\mathbf{M}^2-7\mathbf{M}+35\mathbf{I}=\mathbf{0}$ | M1 | Using Cayley-Hamilton Theorem. Condone omitted $\mathbf{I}$ |
| $\Rightarrow \mathbf{M}^4 = 5\mathbf{M}^3+7\mathbf{M}^2-35\mathbf{M}$ | A1 | Correct expression involving $\mathbf{M}^4$ and non-negative powers of $\mathbf{M}$ |
| $= 5(5\mathbf{M}^2+7\mathbf{M}-35\mathbf{I})+7\mathbf{M}^2-35\mathbf{M}$ | M1 | Substituting for $\mathbf{M}^3$ and obtaining expression in required form |
| $= 32\mathbf{M}^2-175\mathbf{I}$ | A1 | $a=32,\; b=0,\; c=-175$ |3 (i) Show that the characteristic equation of the matrix
$$\mathbf { M } = \left( \begin{array} { r r r }
3 & - 1 & 2 \\
- 4 & 3 & 2 \\
2 & 1 & - 1
\end{array} \right)$$
is $\lambda ^ { 3 } - 5 \lambda ^ { 2 } - 7 \lambda + 35 = 0$.\\
(ii) Show that $\lambda = 5$ is an eigenvalue of $\mathbf { M }$, and find its other eigenvalues.\\
(iii) Find an eigenvector, $\mathbf { v }$, of unit length corresponding to $\lambda = 5$.
State the magnitudes and directions of the vectors $\mathbf { M } ^ { 2 } \mathbf { v }$ and $\mathbf { M } ^ { - 1 } \mathbf { v }$.\\
(iv) Use the Cayley-Hamilton theorem to find the constants $a , b , c$ such that
$$\mathbf { M } ^ { 4 } = a \mathbf { M } ^ { 2 } + b \mathbf { M } + c \mathbf { I } .$$
Section B (18 marks)
\hfill \mbox{\textit{OCR MEI FP2 2012 Q3 [18]}}