| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Eigenvalues and eigenvectors |
| Difficulty | Standard +0.8 This is a comprehensive 3×3 eigenvalue problem from Further Maths covering characteristic equations, eigenvector verification and finding, diagonalization, and Cayley-Hamilton theorem application. While systematic, it requires multiple techniques across several parts, making it moderately challenging but still within standard FP2 scope. The 3×3 characteristic polynomial and Cayley-Hamilton application add computational complexity beyond typical A-level questions. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det(\mathbf{M} - \lambda\mathbf{I}) = (5-\lambda)((-3-\lambda)(4-\lambda)+2)\) | M1A1 | M1 for attempt at \(\det(\mathbf{M}-\lambda\mathbf{I})\) |
| \(+(4(4-\lambda)+4)\) | A1 | A1 each term correct |
| \(+3(4-2(-3-\lambda))\) | A1 | |
| Simplify to \(\lambda^3 - 6\lambda^2 - 7\lambda = 0\) | A1 | A0 if '\(= 0\)' never appears |
| Solve to \(\lambda = -1, 0, 7\) | M1A1 | M1 for eigenvalues are roots of char eqn |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Show that \(\mathbf{M}(1\ \ 3\ \ {-1})^T = (-1\ \ {-3}\ \ 1)^T\) | M1, A1 | For clear evidence of understanding; e.g. just finding eigenvector for \(\lambda=7\) would earn M1A0A0B1B1 |
| Show that \(\mathbf{M}(1\ \ 2\ \ {-1})^T = (0\ \ 0\ \ 0)^T\) | A1 | A1 each calculation |
| Obtain equations \(-2a+3c=1,\ 2a-c=5\) or equivalent | B1 | FT two correct equations |
| Solve to obtain \((4\ \ 1\ \ 3)^T\) | B1 | CAO; Accept \(a=4,\ c=3\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = \begin{pmatrix}1&1&4\\3&2&1\\-1&-1&3\end{pmatrix}\ D = \begin{pmatrix}-1&0&0\\0&0&0\\0&0&7\end{pmatrix}\) | B1B1 | FT; for B2, order must be consistent |
| \(\mathbf{M}^4 = \mathbf{P}\mathbf{D}^4\mathbf{P}^{-1}\) where \(\mathbf{D}^4 = \text{diag}(1\ \ 0\ \ 2401)\) | B1 | CAO |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| C-H: \(\mathbf{M}^3 = 6\mathbf{M}^2 + 7\mathbf{M}\) | M1 | |
| \(\mathbf{M}^4 = 6\mathbf{M}^3 + 7\mathbf{M}^2\) | A1 | CAO |
| \(= 6(6\mathbf{M}^2 + 7\mathbf{M}) + 7\mathbf{M}^2\) | ||
| \(= 43\mathbf{M}^2 + 42\mathbf{M}\) | A1 | CAO |
| [3] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det(\mathbf{M} - \lambda\mathbf{I}) = (5-\lambda)((-3-\lambda)(4-\lambda)+2)$ | M1A1 | M1 for attempt at $\det(\mathbf{M}-\lambda\mathbf{I})$ |
| $+(4(4-\lambda)+4)$ | A1 | A1 each term correct |
| $+3(4-2(-3-\lambda))$ | A1 | |
| Simplify to $\lambda^3 - 6\lambda^2 - 7\lambda = 0$ | A1 | A0 if '$= 0$' never appears |
| Solve to $\lambda = -1, 0, 7$ | M1A1 | M1 for eigenvalues are roots of char eqn |
| | **[7]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Show that $\mathbf{M}(1\ \ 3\ \ {-1})^T = (-1\ \ {-3}\ \ 1)^T$ | M1, A1 | For clear evidence of understanding; e.g. just finding eigenvector for $\lambda=7$ would earn M1A0A0B1B1 |
| Show that $\mathbf{M}(1\ \ 2\ \ {-1})^T = (0\ \ 0\ \ 0)^T$ | A1 | A1 each calculation |
| Obtain equations $-2a+3c=1,\ 2a-c=5$ or equivalent | B1 | FT two correct equations |
| Solve to obtain $(4\ \ 1\ \ 3)^T$ | B1 | CAO; Accept $a=4,\ c=3$ |
| | **[5]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = \begin{pmatrix}1&1&4\\3&2&1\\-1&-1&3\end{pmatrix}\ D = \begin{pmatrix}-1&0&0\\0&0&0\\0&0&7\end{pmatrix}$ | B1B1 | FT; for B2, order must be consistent |
| $\mathbf{M}^4 = \mathbf{P}\mathbf{D}^4\mathbf{P}^{-1}$ where $\mathbf{D}^4 = \text{diag}(1\ \ 0\ \ 2401)$ | B1 | CAO |
| | **[3]** | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| C-H: $\mathbf{M}^3 = 6\mathbf{M}^2 + 7\mathbf{M}$ | M1 | |
| $\mathbf{M}^4 = 6\mathbf{M}^3 + 7\mathbf{M}^2$ | A1 | CAO |
| $= 6(6\mathbf{M}^2 + 7\mathbf{M}) + 7\mathbf{M}^2$ | | |
| $= 43\mathbf{M}^2 + 42\mathbf{M}$ | A1 | CAO |
| | **[3]** | |
---3 This question concerns the matrix $\mathbf { M }$ where $\mathbf { M } = \left( \begin{array} { r r r } 5 & - 1 & 3 \\ 4 & - 3 & - 2 \\ 2 & 1 & 4 \end{array} \right)$.\\
(i) Obtain the characteristic equation of $\mathbf { M }$.
Find the eigenvalues of $\mathbf { M }$.
These eigenvalues are denoted by $\lambda _ { 1 } , \lambda _ { 2 } , \lambda _ { 3 }$, where $\lambda _ { 1 } < \lambda _ { 2 } < \lambda _ { 3 }$.\\
(ii) Verify that an eigenvector corresponding to $\lambda _ { 1 }$ is $\left( \begin{array} { r } 1 \\ 3 \\ - 1 \end{array} \right)$ and that an eigenvector corresponding to $\lambda _ { 2 }$ is $\left( \begin{array} { r } 1 \\ 2 \\ - 1 \end{array} \right)$. Find an eigenvector of the form $\left( \begin{array} { l } a \\ 1 \\ c \end{array} \right)$ corresponding to $\lambda _ { 3 }$.\\
(iii) Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { M } = \mathbf { P D P } ^ { - 1 }$. (You are not required to calculate $\mathbf { P } ^ { - 1 }$.)
Hence write down an expression for $\mathbf { M } ^ { 4 }$ in terms of $\mathbf { P }$ and a diagonal matrix. You should give the elements of the diagonal matrix explicitly.\\
(iv) Use the Cayley-Hamilton theorem to obtain an expression for $\mathbf { M } ^ { 4 }$ as a linear combination of $\mathbf { M }$ and $\mathbf { M } ^ { 2 }$.
\hfill \mbox{\textit{OCR MEI FP2 2015 Q3 [18]}}