| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| 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 Maths question combining characteristic equations with Cayley-Hamilton theorem. Part (a) requires equating coefficients (trace, determinant) which is routine. Part (b) applies the standard Cayley-Hamilton technique to find the inverse by rearranging the characteristic equation. While it's Further Maths content, the method is algorithmic and well-practiced, making it slightly easier than average A-level difficulty overall. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{vmatrix} 1-\lambda & 0 & a \\ -3 & b-\lambda & 1 \\ 0 & 1 & a-\lambda \end{vmatrix} = (1-\lambda)[(b-\lambda)(a-\lambda)-1]+a(-3)(=0)\) | M1 | Correct method to find characteristic equation for M, condone missing \(=0\), one slip allowed if intention is clear |
| \(\lambda^3-(a+b+1)\lambda^2+(a+b+ab-1)\lambda+(3a+1-ab)(=0)\) o.e. | A1 | Multiplies out to achieve correct characteristic equation, condone missing \(=0\) |
| \(\lambda^2 \Rightarrow a+b+1=7\) and \(\lambda \Rightarrow a+b+ab-1=13\); solves simultaneously e.g. \(a+b=6\), \(ab=8\), leading to \(a^2-6a+8=0 \Rightarrow a=\ldots\) | M1 | Complete method to find values of constants \(a\) or \(b\). Equates coefficients for \(\lambda^2\) and \(\lambda\) and solves simultaneously |
| \(a=2,\ b=4\) | A1 | Deduces correct values for \(a\) and \(b\) \((a |
| \(c=-1\) | A1 | Deduces correct value for \(c\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{M}^3-7\mathbf{M}^2+13\mathbf{M}+\text{'their }c\text{'}\mathbf{I}=0\) | B1ft | Uses Cayley-Hamilton theorem replacing \(\lambda\) with M and constant term with constant multiple of I. Follow through on their \(c\). May be implied by M mark |
| \(I=M^3-7M^2+13M \Rightarrow M^{-1}=M^2-7M+13I\) \(\Rightarrow M^{-1}=\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}^2 -7\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}+\begin{pmatrix}13&0&0\\0&13&0\\0&0&13\end{pmatrix}=\ldots\) \(=\begin{pmatrix}1&2&6\\-15&17&0\\-3&6&5\end{pmatrix}-7\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}+\begin{pmatrix}13&0&0\\0&13&0\\0&0&13\end{pmatrix}=\ldots\) | M1 | Complete method to find \(M^{-1}\) using Cayley-Hamilton theorem. Minimum is writing expression for \(M^{-1}\) from characteristic equation e.g. \(M^{-1}=M^2-7M+13I\) |
| \(M^{-1}=\begin{pmatrix}7&2&-8\\6&2&-7\\-3&-1&4\end{pmatrix}\) | A1 | Correct \(M^{-1}\) |
| (3) | ||
| (8 marks) |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 1-\lambda & 0 & a \\ -3 & b-\lambda & 1 \\ 0 & 1 & a-\lambda \end{vmatrix} = (1-\lambda)[(b-\lambda)(a-\lambda)-1]+a(-3)(=0)$ | M1 | Correct method to find characteristic equation for **M**, condone missing $=0$, one slip allowed if intention is clear |
| $\lambda^3-(a+b+1)\lambda^2+(a+b+ab-1)\lambda+(3a+1-ab)(=0)$ o.e. | A1 | Multiplies out to achieve correct characteristic equation, condone missing $=0$ |
| $\lambda^2 \Rightarrow a+b+1=7$ and $\lambda \Rightarrow a+b+ab-1=13$; solves simultaneously e.g. $a+b=6$, $ab=8$, leading to $a^2-6a+8=0 \Rightarrow a=\ldots$ | M1 | Complete method to find values of constants $a$ or $b$. Equates coefficients for $\lambda^2$ and $\lambda$ and solves simultaneously |
| $a=2,\ b=4$ | A1 | Deduces correct values for $a$ and $b$ $(a<b)$ following correct simultaneous equations |
| $c=-1$ | A1 | Deduces correct value for $c$ |
| | **(5)** | |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}^3-7\mathbf{M}^2+13\mathbf{M}+\text{'their }c\text{'}\mathbf{I}=0$ | B1ft | Uses Cayley-Hamilton theorem replacing $\lambda$ with **M** and constant term with constant multiple of **I**. Follow through on their $c$. May be implied by M mark |
| $I=M^3-7M^2+13M \Rightarrow M^{-1}=M^2-7M+13I$ $\Rightarrow M^{-1}=\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}^2 -7\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}+\begin{pmatrix}13&0&0\\0&13&0\\0&0&13\end{pmatrix}=\ldots$ $=\begin{pmatrix}1&2&6\\-15&17&0\\-3&6&5\end{pmatrix}-7\begin{pmatrix}1&0&2\\-3&4&1\\0&1&2\end{pmatrix}+\begin{pmatrix}13&0&0\\0&13&0\\0&0&13\end{pmatrix}=\ldots$ | M1 | Complete method to find $M^{-1}$ using Cayley-Hamilton theorem. Minimum is writing expression for $M^{-1}$ from characteristic equation e.g. $M^{-1}=M^2-7M+13I$ |
| $M^{-1}=\begin{pmatrix}7&2&-8\\6&2&-7\\-3&-1&4\end{pmatrix}$ | A1 | Correct $M^{-1}$ |
| | **(3)** | |
| | **(8 marks)** | |
---\begin{enumerate}
\item Matrix $\mathbf { M }$ is given by
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & 0 & a \\
- 3 & b & 1 \\
0 & 1 & a
\end{array} \right)$$
where $a$ and $b$ are integers, such that $a < b$\\
Given that the characteristic equation for $\mathbf { M }$ is
$$\lambda ^ { 3 } - 7 \lambda ^ { 2 } + 13 \lambda + c = 0$$
where $c$ is a constant,\\
(a) determine the values of $a , b$ and $c$.\\
(b) Hence, using the Cayley-Hamilton theorem, determine the matrix $\mathbf { M } ^ { - 1 }$
\hfill \mbox{\textit{Edexcel FP2 2022 Q2 [8]}}