| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| 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 straightforward Further Maths question testing standard techniques: finding a characteristic equation (routine determinant expansion), using det(M) to find k (simple substitution), and applying Cayley-Hamilton to find an inverse (direct application of a learned method). While it involves matrices and Further Maths content, each part follows a well-practiced algorithm with no novel insight required, making it slightly easier than average for FP2. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sight of \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\) | B1 | Recalls characteristic equation is found using \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\) |
| Attempts to expand \(\begin{vmatrix} 1-\lambda & k & -2 \\ 2 & -4-\lambda & 1 \\ 1 & 2 & 3-\lambda \end{vmatrix} = 0\) giving \((1-\lambda)[(-4-\lambda)(3-\lambda)-2] - k[2(3-\lambda)-1] + (-2)[4-(-4-\lambda)] = 0\) | M1 | Attempts to expand the determinant |
| \(\Rightarrow (1-\lambda)(\lambda^2 + \lambda - 14) - k(5-2\lambda) - 16 - 2\lambda = 0\) \(\Rightarrow \lambda^3 - (2k+13)\lambda + 5(k+6) = 0\)* | A1\* | Achieves correct equation with no errors and at least one intermediate step following the expansion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm 5(k+6) = 5 \Rightarrow k = \ldots\) or \((-12-2) - k(6-1) - 2(4+4) = 5 \Rightarrow k = \ldots\) | M1 | Attempts to use determinant equals 5 to find \(k\). May use original matrix or \(-5(k+6)\) from part (a) (allow \(\pm\)) |
| \(k = -7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Hence by C-H theorem \(\mathbf{M}^3 + \mathbf{M} - 5\mathbf{I} = \mathbf{0}\) | M1 | Attempts to use Cayley-Hamilton theorem; equation correct for their \(k\), with correct use of \(\mathbf{I}\) |
| Multiplying by \(\mathbf{M}^{-1}\) gives \(\mathbf{M}^2 + \mathbf{I} - 5\mathbf{M}^{-1} = \mathbf{0} \Rightarrow \mathbf{M}^{-1} = \ldots\) | M1 | Realises need to multiply through by \(\mathbf{M}^{-1}\) and rearrange |
| \(\mathbf{M}^{-1} = \frac{1}{5}(\mathbf{M}^2 + \mathbf{I})\) | A1 | |
| \(= \frac{1}{5}\left[\begin{pmatrix}-15 & 17 & -15 \\ -5 & 4 & -5 \\ 8 & -9 & 9\end{pmatrix} + \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\right] = \ldots\) | M1 | Proceeds to find \(\mathbf{M}^{-1}\) from their equation |
| \(= \frac{1}{5}\begin{pmatrix}-14 & 17 & -15 \\ -5 & 5 & -5 \\ 8 & -9 & 10\end{pmatrix}\) or \(\begin{pmatrix}-\frac{14}{5} & \frac{17}{5} & -3 \\ -1 & 1 & -1 \\ \frac{8}{5} & -\frac{9}{5} & 2\end{pmatrix}\) | A1 | Correct answer |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sight of $\det(\mathbf{M} - \lambda\mathbf{I}) = 0$ | **B1** | Recalls characteristic equation is found using $\det(\mathbf{M} - \lambda\mathbf{I}) = 0$ |
| Attempts to expand $\begin{vmatrix} 1-\lambda & k & -2 \\ 2 & -4-\lambda & 1 \\ 1 & 2 & 3-\lambda \end{vmatrix} = 0$ giving $(1-\lambda)[(-4-\lambda)(3-\lambda)-2] - k[2(3-\lambda)-1] + (-2)[4-(-4-\lambda)] = 0$ | **M1** | Attempts to expand the determinant |
| $\Rightarrow (1-\lambda)(\lambda^2 + \lambda - 14) - k(5-2\lambda) - 16 - 2\lambda = 0$ $\Rightarrow \lambda^3 - (2k+13)\lambda + 5(k+6) = 0$* | **A1\*** | Achieves correct equation with no errors and at least one intermediate step following the expansion |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm 5(k+6) = 5 \Rightarrow k = \ldots$ or $(-12-2) - k(6-1) - 2(4+4) = 5 \Rightarrow k = \ldots$ | **M1** | Attempts to use determinant equals 5 to find $k$. May use original matrix or $-5(k+6)$ from part (a) (allow $\pm$) |
| $k = -7$ | **A1** | |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Hence by C-H theorem $\mathbf{M}^3 + \mathbf{M} - 5\mathbf{I} = \mathbf{0}$ | **M1** | Attempts to use Cayley-Hamilton theorem; equation correct for their $k$, with correct use of $\mathbf{I}$ |
| Multiplying by $\mathbf{M}^{-1}$ gives $\mathbf{M}^2 + \mathbf{I} - 5\mathbf{M}^{-1} = \mathbf{0} \Rightarrow \mathbf{M}^{-1} = \ldots$ | **M1** | Realises need to multiply through by $\mathbf{M}^{-1}$ and rearrange |
| $\mathbf{M}^{-1} = \frac{1}{5}(\mathbf{M}^2 + \mathbf{I})$ | **A1** | |
| $= \frac{1}{5}\left[\begin{pmatrix}-15 & 17 & -15 \\ -5 & 4 & -5 \\ 8 & -9 & 9\end{pmatrix} + \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\right] = \ldots$ | **M1** | Proceeds to find $\mathbf{M}^{-1}$ from their equation |
| $= \frac{1}{5}\begin{pmatrix}-14 & 17 & -15 \\ -5 & 5 & -5 \\ 8 & -9 & 10\end{pmatrix}$ or $\begin{pmatrix}-\frac{14}{5} & \frac{17}{5} & -3 \\ -1 & 1 & -1 \\ \frac{8}{5} & -\frac{9}{5} & 2\end{pmatrix}$ | **A1** | Correct answer |
---3.
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & k & - 2 \\
2 & - 4 & 1 \\
1 & 2 & 3
\end{array} \right)$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that, in terms of $k$, a characteristic equation for $\mathbf { M }$ is given by
$$\lambda ^ { 3 } - ( 2 k + 13 ) \lambda + 5 ( k + 6 ) = 0$$
Given that $\operatorname { det } \mathbf { M } = 5$
\item \begin{enumerate}[label=(\roman*)]
\item find the value of $k$
\item use the Cayley-Hamilton theorem to find the inverse of $\mathbf { M }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2020 Q3 [10]}}