| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Use Cayley-Hamilton for matrix power |
| Difficulty | Standard +0.3 This is a straightforward application of the Cayley-Hamilton theorem requiring students to find a characteristic equation, substitute the matrix into it, then manipulate to match a given form. While it involves Further Maths content, the steps are mechanical and follow a standard template with no novel insight required, making it slightly easier than average. |
| Spec | 4.03h Determinant 2x2: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\det\begin{pmatrix}-1-\lambda & a \\ 3 & 8-\lambda\end{pmatrix} = 0 \Rightarrow (-1-\lambda)(8-\lambda) - 3a = 0 \Rightarrow \ldots\) | M1 | 1.1b |
| \(\lambda^2 - 7\lambda - 3a - 8 = 0\) o.e. | A1 | 1.1b |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{A}^2 - 7\mathbf{A} - 3a\mathbf{I} - 8\mathbf{I} = 0 \Rightarrow \mathbf{A}^3 = 7\mathbf{A}^2 + (3a+8)\mathbf{A}\) | M1 | 1.1b |
| \(\Rightarrow \mathbf{A}^3 = 7\big(7\mathbf{A} + (3a+8)\mathbf{I}\big) + (3a+8)\mathbf{A}\), Or direct matrix computation expanding \(\mathbf{A}^3 = 7\begin{pmatrix}-1 & a\\3 & 8\end{pmatrix}^2 + (8+3a)\begin{pmatrix}-1 & a\\3 & 8\end{pmatrix}\) giving \(= 7\begin{pmatrix}1+3a & 7a\\21 & 3a+64\end{pmatrix} + \begin{pmatrix}-8-3a & 8a+3a^2\\24+9a & 64+24a\end{pmatrix}\) | M1 | 2.1 |
| \(\Rightarrow \mathbf{A}^3 = (3a+8+49)\mathbf{A} + 7(3a+8)\mathbf{I} \Rightarrow 3a + 57 = 1 \Rightarrow a = \ldots\) Or \(\begin{pmatrix}b-1 & a\\3 & b+8\end{pmatrix} = \begin{pmatrix}18a-1 & 3a^2+57a\\171+9a & 512+27a\end{pmatrix} \Rightarrow\) e.g. \(3 = 171 + 9a \Rightarrow a = \ldots\) Or \(\mathbf{A}^3 = \begin{pmatrix}-1 & a\\3 & 8\end{pmatrix} + \begin{pmatrix}18a & 3a^2+56a\\168+9a & 504+27a\end{pmatrix} \Rightarrow\) e.g. \(18a = 504 + 27a \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(\Rightarrow a = -\dfrac{56}{3},\ b = -336\) | A1 | 1.1b |
| (4) | ||
| (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct method to find the characteristic equation for A | M1 | Condone missing \(= 0\) |
| Correct simplified characteristic equation | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses Cayley-Hamilton theorem with their equation and multiplies through by A to find equation for \(\mathbf{A}^3\) | M1 | Must use Cayley-Hamilton; any other approach scores no marks |
| Substitutes for \(\mathbf{A}^2\) to obtain equation for \(\mathbf{A}^3\) in terms of \(a\), I and A; alternatively substitutes in matrix A and attempts to square | M1 | — |
| Equates coefficient(s) of A to 1 and finds value for \(a\); alternatively equates elements to find \(a\) | M1 | — |
| Correct values for \(a\) and \(b\) | A1 | Special case: missing I scores max M1M1A0 unless I implied |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det\begin{pmatrix}-1-\lambda & a \\ 3 & 8-\lambda\end{pmatrix} = 0 \Rightarrow (-1-\lambda)(8-\lambda) - 3a = 0 \Rightarrow \ldots$ | M1 | 1.1b |
| $\lambda^2 - 7\lambda - 3a - 8 = 0$ o.e. | A1 | 1.1b |
| | **(2)** | |
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{A}^2 - 7\mathbf{A} - 3a\mathbf{I} - 8\mathbf{I} = 0 \Rightarrow \mathbf{A}^3 = 7\mathbf{A}^2 + (3a+8)\mathbf{A}$ | M1 | 1.1b |
| $\Rightarrow \mathbf{A}^3 = 7\big(7\mathbf{A} + (3a+8)\mathbf{I}\big) + (3a+8)\mathbf{A}$, **Or** direct matrix computation expanding $\mathbf{A}^3 = 7\begin{pmatrix}-1 & a\\3 & 8\end{pmatrix}^2 + (8+3a)\begin{pmatrix}-1 & a\\3 & 8\end{pmatrix}$ giving $= 7\begin{pmatrix}1+3a & 7a\\21 & 3a+64\end{pmatrix} + \begin{pmatrix}-8-3a & 8a+3a^2\\24+9a & 64+24a\end{pmatrix}$ | M1 | 2.1 |
| $\Rightarrow \mathbf{A}^3 = (3a+8+49)\mathbf{A} + 7(3a+8)\mathbf{I} \Rightarrow 3a + 57 = 1 \Rightarrow a = \ldots$ **Or** $\begin{pmatrix}b-1 & a\\3 & b+8\end{pmatrix} = \begin{pmatrix}18a-1 & 3a^2+57a\\171+9a & 512+27a\end{pmatrix} \Rightarrow$ e.g. $3 = 171 + 9a \Rightarrow a = \ldots$ **Or** $\mathbf{A}^3 = \begin{pmatrix}-1 & a\\3 & 8\end{pmatrix} + \begin{pmatrix}18a & 3a^2+56a\\168+9a & 504+27a\end{pmatrix} \Rightarrow$ e.g. $18a = 504 + 27a \Rightarrow a = \ldots$ | M1 | 1.1b |
| $\Rightarrow a = -\dfrac{56}{3},\ b = -336$ | A1 | 1.1b |
| | **(4)** | |
| | **(6 marks)** | |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method to find the characteristic equation for **A** | M1 | Condone missing $= 0$ |
| Correct simplified characteristic equation | A1 | — |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses Cayley-Hamilton theorem with their equation and multiplies through by **A** to find equation for $\mathbf{A}^3$ | M1 | Must use Cayley-Hamilton; any other approach scores no marks |
| Substitutes for $\mathbf{A}^2$ to obtain equation for $\mathbf{A}^3$ in terms of $a$, **I** and **A**; alternatively substitutes in matrix **A** and attempts to square | M1 | — |
| Equates coefficient(s) of **A** to 1 and finds value for $a$; alternatively equates elements to find $a$ | M1 | — |
| Correct values for $a$ and $b$ | A1 | Special case: missing **I** scores max M1M1A0 unless **I** implied |
---1.
$$\mathbf { A } = \left( \begin{array} { r r }
- 1 & a \\
3 & 8
\end{array} \right)$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Determine, in expanded form in terms of $a$, the characteristic equation for $\mathbf { A }$.
\item Hence use the Cayley-Hamilton theorem to determine values of $a$ and $b$ such that
$$\mathbf { A } ^ { 3 } = \mathbf { A } + b \mathbf { I }$$
where $\mathbf { I }$ is the $2 \times 2$ identity matrix.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2023 Q1 [6]}}