| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| 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 matrix power |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard Cayley-Hamilton theorem application and matrix reconstruction from eigenvalues/eigenvectors. Part (b) requires simple algebraic manipulation (multiplying the characteristic equation by A), and part (ii) uses the standard formula M = PDP^(-1). All techniques are routine for FP2 students with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{vmatrix} 1-\lambda & -2 \\ 1 & 4-\lambda \end{vmatrix} = (1-\lambda)(4-\lambda) + 2 = 0\) | M1 | Attempts the determinant of \(\mathbf{A} - \lambda\mathbf{I}\) |
| \(\Rightarrow 4 - 5\lambda + \lambda^2 + 2 = 0 \Rightarrow \lambda^2 - 5\lambda + 6 = 0\) | A1* | Fully correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{A}^2 - 5\mathbf{A} + 6\mathbf{I} = 0\) | M1 | Applies the Cayley-Hamilton theorem to equation from (a)(i) |
| \(\mathbf{A}^3 - 5\mathbf{A}^2 + 6\mathbf{A} = 0 \Rightarrow \mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}\) | M1 | Full method leading to \(\mathbf{A}^3\) by multiplying by \(\mathbf{A}\) and substituting for \(\mathbf{A}^2\) |
| \(\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}\) | A1 | Deduces correct expression or correct values for \(p\) and \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda^2 - 5\lambda + 6 = 0 \Rightarrow \lambda^3 - 5\lambda^2 + 6\lambda = 0 \Rightarrow \lambda^3 = 5(5\lambda - 6) - 6\lambda\) | M1 | Full method leading to \(\lambda^3\) in terms of \(\lambda\) |
| \(\mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}\) | M1 | Applies the Cayley-Hamilton theorem |
| \(\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}\) | A1 | Deduces correct values for \(p\) and \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2-\text{i} \end{pmatrix} = (-1+\text{i})\begin{pmatrix} 1 \\ 2-\text{i} \end{pmatrix}\) or conjugate equation | M1 | Uses a general matrix and sets up at least one matrix equation using information given |
| \(a + b(2-\text{i}) = -1+\text{i}\), \(a + b(2+\text{i}) = -1-\text{i}\), \(c + d(2-\text{i}) = -1+3\text{i}\), \(c + d(2+\text{i}) = -1-3\text{i}\) | A1 | Correct equations in terms of \(a\), \(b\), \(c\) and \(d\) |
| Solving simultaneously: \(a = 1\), \(b = -1\) | M1, A1 | Solves simultaneously; one correct pair of values |
| \(c = 5\), \(d = -3\) | ||
| \(\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}\) | A1 | Deduces the correct matrix \(\mathbf{M}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2-\text{i} & 2+\text{i} \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \frac{1}{2\text{i}}\begin{pmatrix} 2+\text{i} & -1 \\ \text{i}-2 & 1 \end{pmatrix}\) | M1, A1 | Attempts to find the inverse of the matrix of eigenvectors; correct matrix |
| \(\mathbf{D} = \mathbf{P}^{-1}\mathbf{M}\mathbf{P} \Rightarrow \mathbf{M} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), \(\mathbf{M} = \frac{1}{2\text{i}}\begin{pmatrix} 1 & 1 \\ 2-\text{i} & 2+\text{i} \end{pmatrix}\begin{pmatrix} -1+\text{i} & 0 \\ 0 & -1-\text{i} \end{pmatrix}\begin{pmatrix} 2+\text{i} & -1 \\ \text{i}-2 & 1 \end{pmatrix}\) | M1 | Attempts \(\mathbf{PDP}^{-1}\) where \(\mathbf{D}\) is the diagonal matrix of eigenvalues |
| \(\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}\) | A1, A1 | At least 2 elements correct; deduces the correct matrix \(\mathbf{M}\) |
## Question 3(i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 1-\lambda & -2 \\ 1 & 4-\lambda \end{vmatrix} = (1-\lambda)(4-\lambda) + 2 = 0$ | M1 | Attempts the determinant of $\mathbf{A} - \lambda\mathbf{I}$ |
| $\Rightarrow 4 - 5\lambda + \lambda^2 + 2 = 0 \Rightarrow \lambda^2 - 5\lambda + 6 = 0$ | A1* | Fully correct proof |
## Question 3(i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{A}^2 - 5\mathbf{A} + 6\mathbf{I} = 0$ | M1 | Applies the Cayley-Hamilton theorem to equation from (a)(i) |
| $\mathbf{A}^3 - 5\mathbf{A}^2 + 6\mathbf{A} = 0 \Rightarrow \mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}$ | M1 | Full method leading to $\mathbf{A}^3$ by multiplying by $\mathbf{A}$ and substituting for $\mathbf{A}^2$ |
| $\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}$ | A1 | Deduces correct expression or correct values for $p$ and $q$ |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda^2 - 5\lambda + 6 = 0 \Rightarrow \lambda^3 - 5\lambda^2 + 6\lambda = 0 \Rightarrow \lambda^3 = 5(5\lambda - 6) - 6\lambda$ | M1 | Full method leading to $\lambda^3$ in terms of $\lambda$ |
| $\mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}$ | M1 | Applies the Cayley-Hamilton theorem |
| $\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}$ | A1 | Deduces correct values for $p$ and $q$ |
## Question 3(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2-\text{i} \end{pmatrix} = (-1+\text{i})\begin{pmatrix} 1 \\ 2-\text{i} \end{pmatrix}$ or conjugate equation | M1 | Uses a general matrix and sets up at least one matrix equation using information given |
| $a + b(2-\text{i}) = -1+\text{i}$, $a + b(2+\text{i}) = -1-\text{i}$, $c + d(2-\text{i}) = -1+3\text{i}$, $c + d(2+\text{i}) = -1-3\text{i}$ | A1 | Correct equations in terms of $a$, $b$, $c$ and $d$ |
| Solving simultaneously: $a = 1$, $b = -1$ | M1, A1 | Solves simultaneously; one correct pair of values |
| $c = 5$, $d = -3$ | | |
| $\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}$ | A1 | Deduces the correct matrix $\mathbf{M}$ |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2-\text{i} & 2+\text{i} \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \frac{1}{2\text{i}}\begin{pmatrix} 2+\text{i} & -1 \\ \text{i}-2 & 1 \end{pmatrix}$ | M1, A1 | Attempts to find the inverse of the matrix of eigenvectors; correct matrix |
| $\mathbf{D} = \mathbf{P}^{-1}\mathbf{M}\mathbf{P} \Rightarrow \mathbf{M} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, $\mathbf{M} = \frac{1}{2\text{i}}\begin{pmatrix} 1 & 1 \\ 2-\text{i} & 2+\text{i} \end{pmatrix}\begin{pmatrix} -1+\text{i} & 0 \\ 0 & -1-\text{i} \end{pmatrix}\begin{pmatrix} 2+\text{i} & -1 \\ \text{i}-2 & 1 \end{pmatrix}$ | M1 | Attempts $\mathbf{PDP}^{-1}$ where $\mathbf{D}$ is the diagonal matrix of eigenvalues |
| $\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}$ | A1, A1 | At least 2 elements correct; deduces the correct matrix $\mathbf{M}$ |
---\begin{enumerate}
\item (i)
\end{enumerate}
$$A = \left( \begin{array} { r r }
1 & - 2 \\
1 & 4
\end{array} \right)$$
(a) Show that the characteristic equation for $\mathbf { A }$ is $\lambda ^ { 2 } - 5 \lambda + 6 = 0$\\
(b) Use the Cayley-Hamilton theorem to find integers $p$ and $q$ such that
$$\mathbf { A } ^ { 3 } = p \mathbf { A } + q \mathbf { I }$$
(ii) Given that the $2 \times 2$ matrix $\mathbf { M }$ has eigenvalues $- 1 + \mathrm { i }$ and $- 1 - \mathrm { i }$, with eigenvectors $\binom { 1 } { 2 - \mathrm { i } }$ and $\binom { 1 } { 2 + \mathrm { i } }$ respectively, find the matrix $\mathbf { M }$.
\hfill \mbox{\textit{Edexcel FP2 AS 2020 Q3 [10]}}