| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Use Cayley-Hamilton for inverse |
| Difficulty | Moderate -0.3 This is a straightforward application of the Cayley-Hamilton theorem to find a matrix inverse. Part (a) requires computing a 2×2 characteristic equation (routine), and part (b) involves algebraic manipulation of the Cayley-Hamilton result to isolate A^{-1}. While this is a Further Maths topic, the execution is mechanical with no conceptual challenges or novel problem-solving required, making it slightly easier than an average A-level question. |
| Spec | 4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\det\begin{pmatrix}3-\lambda & 2 \\ 2 & 2-\lambda\end{pmatrix} = (3-\lambda)(2-\lambda)-4(=0)\) | M1 | Complete method to find characteristic equation, condone missing \(=0\) |
| \(\lambda^2 - 5\lambda + 2 = 0\) | A1 | Obtains correct three term quadratic – may use any variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{A}^2 - 5\mathbf{A} + 2\mathbf{I} = 0\) | B1ft | Uses Cayley-Hamilton Theorem replacing \(\lambda\) with \(\mathbf{A}\) and constant term with constant multiple of identity \(\mathbf{I}\) |
| Multiplies through by \(\mathbf{A}^{-1}\): \(\mathbf{A} - 5\mathbf{I} + 2\mathbf{A}^{-1} = 0\) and rearranges; OR rearranges to make \(\mathbf{I}\) subject: \(\mathbf{I} = \frac{(5\mathbf{A}-\mathbf{A}^2)}{2} = \mathbf{A}\frac{(5\mathbf{I}-\mathbf{A})}{2} \Rightarrow \mathbf{A}^{-1} = \ldots\); OR \(\mathbf{I} = \frac{5}{2}\mathbf{A} - \frac{1}{2}\mathbf{A}^2 \Rightarrow \mathbf{A}^{-1} = \frac{5}{2}\mathbf{A}\mathbf{A}^{-1} - \frac{1}{2}\mathbf{A}^2\mathbf{A}^{-1}\) | M1 | Complete method using part (a) to find \(\mathbf{A}^{-1}\) |
| \(\mathbf{A}^{-1} = -\frac{1}{2}\mathbf{A} + \frac{5}{2}\mathbf{I}\) | A1 | Correct expression for \(\mathbf{A}^{-1}\), must use answer to part (a) |
## Question 1:
**Part (a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det\begin{pmatrix}3-\lambda & 2 \\ 2 & 2-\lambda\end{pmatrix} = (3-\lambda)(2-\lambda)-4(=0)$ | M1 | Complete method to find characteristic equation, condone missing $=0$ |
| $\lambda^2 - 5\lambda + 2 = 0$ | A1 | Obtains correct three term quadratic – may use any variable |
**Part (b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{A}^2 - 5\mathbf{A} + 2\mathbf{I} = 0$ | B1ft | Uses Cayley-Hamilton Theorem replacing $\lambda$ with $\mathbf{A}$ and constant term with constant multiple of identity $\mathbf{I}$ |
| Multiplies through by $\mathbf{A}^{-1}$: $\mathbf{A} - 5\mathbf{I} + 2\mathbf{A}^{-1} = 0$ and rearranges; OR rearranges to make $\mathbf{I}$ subject: $\mathbf{I} = \frac{(5\mathbf{A}-\mathbf{A}^2)}{2} = \mathbf{A}\frac{(5\mathbf{I}-\mathbf{A})}{2} \Rightarrow \mathbf{A}^{-1} = \ldots$; OR $\mathbf{I} = \frac{5}{2}\mathbf{A} - \frac{1}{2}\mathbf{A}^2 \Rightarrow \mathbf{A}^{-1} = \frac{5}{2}\mathbf{A}\mathbf{A}^{-1} - \frac{1}{2}\mathbf{A}^2\mathbf{A}^{-1}$ | M1 | Complete method using part (a) to find $\mathbf{A}^{-1}$ |
| $\mathbf{A}^{-1} = -\frac{1}{2}\mathbf{A} + \frac{5}{2}\mathbf{I}$ | A1 | Correct expression for $\mathbf{A}^{-1}$, must use answer to part (a) |
---\begin{enumerate}
\item Given that
\end{enumerate}
$$\mathbf { A } = \left( \begin{array} { l l }
3 & 2 \\
2 & 2
\end{array} \right)$$
(a) find the characteristic equation for the matrix $\mathbf { A }$, simplifying your answer.\\
(b) Hence find an expression for the matrix $\mathbf { A } ^ { - 1 }$ in the form $\lambda \mathbf { A } + \mu \mathbf { I }$, where $\lambda$ and $\mu$ are constants to be found.
\hfill \mbox{\textit{Edexcel FP2 AS 2019 Q1 [5]}}