| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues/vectors of matrix combination |
| Difficulty | Standard +0.3 Part (i) is a straightforward proof using the definition of eigenvectors (Ae = λe implies A³e = λ³e). Part (ii) requires finding eigenvalues and eigenvectors of the given matrix A, then applying the result to A³ + I, which involves routine diagonalization techniques. This is standard Further Maths material with no novel insights required, making it slightly easier than average. |
| Spec | 4.03a Matrix language: terminology and notation4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{Ae} = \lambda\mathbf{e}\) SOI | B1 | |
| \(\mathbf{A}^3\mathbf{e} = \mathbf{A}^2(\mathbf{Ae}) = \lambda\mathbf{A}(\mathbf{Ae}) = \lambda^2(\mathbf{Ae}) = \lambda^3\mathbf{e}\) | M1 | Substitutes for \(\mathbf{Ae}\) |
| So eigenvalue is \(\lambda^3\). Special case: states eigenvalue is \(\lambda^3\) | A1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of \(\mathbf{A}\) are 2 and 3 | B1 | |
| Eigenvectors of \(\mathbf{A}\) are \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\1\end{pmatrix}\) | M1 | AEF (allow any non-zero scalar multiple) |
| So \(\mathbf{P} = \begin{pmatrix}1 & 0\\1 & 1\end{pmatrix}\) | A1 | Alt method: Find \(\mathbf{A}^3 + \mathbf{I} = \begin{pmatrix}9 & 0\\-19 & 28\end{pmatrix}\) B1 |
| and \(\mathbf{D} = \begin{pmatrix}2^3+1 & 0\\0 & 3^3+1\end{pmatrix} = \begin{pmatrix}9 & 0\\0 & 28\end{pmatrix}\) | M1 A1FT | Eigenvalues \((9, 28)\) and vectors \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\1\end{pmatrix}\) M1, A1; \(\mathbf{P}\) and \(\mathbf{D}\) FT on eigenvalues M1, A1FT |
| Columns of \(\mathbf{P}\) and \(\mathbf{D}\) can be permuted, but must match |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{Ae} = \lambda\mathbf{e}$ SOI | B1 | |
| $\mathbf{A}^3\mathbf{e} = \mathbf{A}^2(\mathbf{Ae}) = \lambda\mathbf{A}(\mathbf{Ae}) = \lambda^2(\mathbf{Ae}) = \lambda^3\mathbf{e}$ | M1 | Substitutes for $\mathbf{Ae}$ |
| So eigenvalue is $\lambda^3$. Special case: states eigenvalue is $\lambda^3$ | A1 B1 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of $\mathbf{A}$ are 2 and 3 | B1 | |
| Eigenvectors of $\mathbf{A}$ are $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ | M1 | AEF (allow any non-zero scalar multiple) |
| So $\mathbf{P} = \begin{pmatrix}1 & 0\\1 & 1\end{pmatrix}$ | A1 | Alt method: Find $\mathbf{A}^3 + \mathbf{I} = \begin{pmatrix}9 & 0\\-19 & 28\end{pmatrix}$ B1 |
| and $\mathbf{D} = \begin{pmatrix}2^3+1 & 0\\0 & 3^3+1\end{pmatrix} = \begin{pmatrix}9 & 0\\0 & 28\end{pmatrix}$ | M1 A1FT | Eigenvalues $(9, 28)$ and vectors $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ M1, A1; $\mathbf{P}$ and $\mathbf{D}$ FT on eigenvalues M1, A1FT |
| Columns of $\mathbf{P}$ and $\mathbf{D}$ can be permuted, but must match | | |
---5 It is given that $\mathbf { e }$ is an eigenvector of the matrix $\mathbf { A }$ with corresponding eigenvalue $\lambda$.\\
(i) Show that $\mathbf { e }$ is an eigenvector of $\mathbf { A } ^ { 3 }$ and state the corresponding eigenvalue.\\
It is given that
$$\mathbf { A } = \left( \begin{array} { r r }
2 & 0 \\
- 1 & 3
\end{array} \right) .$$
(ii) Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that
$$\mathbf { A } ^ { 3 } + \mathbf { I } = \mathbf { P } \mathbf { D } \mathbf { P } ^ { - 1 }$$
where $\mathbf { I }$ is the $2 \times 2$ identity matrix.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q5 [8]}}