| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹ |
| Difficulty | Challenging +1.2 This is a comprehensive but highly structured 3×3 eigenvalue problem with significant scaffolding. Parts (i)-(iv) guide students through standard procedures (characteristic equation given, one eigenvalue provided, eigenvectors verified rather than found independently). Part (v) is routine matrix construction, and part (vi) applies Cayley-Hamilton in a standard way. While 3×3 matrices involve more computation than 2×2, the extensive guidance and verification steps make this easier than an unscaffolded eigenvalue problem, placing it moderately above average difficulty. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03m det(AB) = det(A)*det(B) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\) expanded correctly | M1 | |
| Correct expansion giving \(\lambda^3 + 6\lambda^2 - 9\lambda - 14 = 0\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substituting \(\lambda = -1\): \((-1)^3 + 6(1) - 9(-1) - 14 = -1+6+9-14 = 0\) | B1 | Verification |
| \((\lambda+1)(\lambda^2 + 5\lambda - 14) = 0\) | M1 | Factorising |
| \((\lambda+1)(\lambda+7)(\lambda-2) = 0\), eigenvalues \(2\) and \(-7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\mathbf{M}+\mathbf{I})\mathbf{x} = 0\): \((2\mathbf{I}+\mathbf{M})\mathbf{x}=0\) | M1 | Setting up equations |
| Equations solved correctly | A1 | |
| Eigenvector e.g. \(\begin{pmatrix}3\\-6\\2\end{pmatrix}\) or equivalent | A1 | Accept any scalar multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}\begin{pmatrix}3\\0\\1\end{pmatrix}\): showing result \(= 2\begin{pmatrix}3\\0\\1\end{pmatrix}\) | M1 A1 | |
| \(\mathbf{M}\begin{pmatrix}0\\3\\-2\end{pmatrix}\): showing result \(= -7\begin{pmatrix}0\\3\\-2\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P} = \begin{pmatrix}3&0&e_1\\0&3&e_2\\1&-2&e_3\end{pmatrix}\) with eigenvectors as columns (any order) | B1 | |
| \(\mathbf{D} = \begin{pmatrix}8&0&0\\0&-343&0\\0&0&-1\end{pmatrix}\) (i.e. \(\lambda^3\) values, matching order) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Cayley-Hamilton: \(\mathbf{M}^3 + 6\mathbf{M}^2 - 9\mathbf{M} - 14\mathbf{I} = 0\) | B1 | |
| Rearranging: \(14\mathbf{M}^{-1} = \mathbf{M}^2 + 6\mathbf{M} - 9\mathbf{I}\) | M1 | Multiplying by \(\mathbf{M}^{-1}\) |
| \(\mathbf{M}^{-1} = \frac{1}{14}\mathbf{M}^2 + \frac{6}{14}\mathbf{M} - \frac{9}{14}\mathbf{I}\) | A1 | i.e. \(a=\frac{1}{14}\), \(b=\frac{3}{7}\), \(c=-\frac{9}{14}\) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det(\mathbf{M} - \lambda\mathbf{I}) = 0$ expanded correctly | M1 | |
| Correct expansion giving $\lambda^3 + 6\lambda^2 - 9\lambda - 14 = 0$ | A1 A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substituting $\lambda = -1$: $(-1)^3 + 6(1) - 9(-1) - 14 = -1+6+9-14 = 0$ | B1 | Verification |
| $(\lambda+1)(\lambda^2 + 5\lambda - 14) = 0$ | M1 | Factorising |
| $(\lambda+1)(\lambda+7)(\lambda-2) = 0$, eigenvalues $2$ and $-7$ | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\mathbf{M}+\mathbf{I})\mathbf{x} = 0$: $(2\mathbf{I}+\mathbf{M})\mathbf{x}=0$ | M1 | Setting up equations |
| Equations solved correctly | A1 | |
| Eigenvector e.g. $\begin{pmatrix}3\\-6\\2\end{pmatrix}$ or equivalent | A1 | Accept any scalar multiple |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}\begin{pmatrix}3\\0\\1\end{pmatrix}$: showing result $= 2\begin{pmatrix}3\\0\\1\end{pmatrix}$ | M1 A1 | |
| $\mathbf{M}\begin{pmatrix}0\\3\\-2\end{pmatrix}$: showing result $= -7\begin{pmatrix}0\\3\\-2\end{pmatrix}$ | A1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P} = \begin{pmatrix}3&0&e_1\\0&3&e_2\\1&-2&e_3\end{pmatrix}$ with eigenvectors as columns (any order) | B1 | |
| $\mathbf{D} = \begin{pmatrix}8&0&0\\0&-343&0\\0&0&-1\end{pmatrix}$ (i.e. $\lambda^3$ values, matching order) | B1 B1 | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley-Hamilton: $\mathbf{M}^3 + 6\mathbf{M}^2 - 9\mathbf{M} - 14\mathbf{I} = 0$ | B1 | |
| Rearranging: $14\mathbf{M}^{-1} = \mathbf{M}^2 + 6\mathbf{M} - 9\mathbf{I}$ | M1 | Multiplying by $\mathbf{M}^{-1}$ |
| $\mathbf{M}^{-1} = \frac{1}{14}\mathbf{M}^2 + \frac{6}{14}\mathbf{M} - \frac{9}{14}\mathbf{I}$ | A1 | i.e. $a=\frac{1}{14}$, $b=\frac{3}{7}$, $c=-\frac{9}{14}$ |
---3 The matrix $\mathbf { M } = \left( \begin{array} { r r r } 1 & 2 & 3 \\ - 2 & - 3 & 6 \\ 2 & 2 & - 4 \end{array} \right)$.\\
(i) Show that the characteristic equation for $\mathbf { M }$ is $\lambda ^ { 3 } + 6 \lambda ^ { 2 } - 9 \lambda - 14 = 0$.\\
(ii) Show that - 1 is an eigenvalue of $\mathbf { M }$, and find the other two eigenvalues.\\
(iii) Find an eigenvector corresponding to the eigenvalue - 1 .\\
(iv) Verify that $\left( \begin{array} { l } 3 \\ 0 \\ 1 \end{array} \right)$ and $\left( \begin{array} { r } 0 \\ 3 \\ - 2 \end{array} \right)$ are eigenvectors of $\mathbf { M }$.\\
(v) Write down a matrix $\mathbf { P }$, and a diagonal matrix $\mathbf { D }$, such that $\mathbf { M } ^ { 3 } = \mathbf { P D P } \mathbf { P } ^ { - 1 }$.\\
(vi) Use the Cayley-Hamilton theorem to express $\mathbf { M } ^ { - 1 }$ in the form $a \mathbf { M } ^ { 2 } + b \mathbf { M } + c \mathbf { I }$.
Section B (18 marks)
\hfill \mbox{\textit{OCR MEI FP2 2006 Q3 [18]}}