| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues/vectors of matrix combination |
| Difficulty | Standard +0.8 Part (i) is a straightforward proof requiring basic manipulation of the eigenvector definition. Part (ii) requires finding eigenvalues of (A+nI)² using the result from (i), then finding eigenvectors and constructing the diagonalization—this involves multiple steps with upper triangular matrices but follows standard procedures. The question is moderately challenging for Further Maths, requiring careful algebraic manipulation and understanding of eigenvalue theory, but doesn't require novel insights beyond applying learned techniques. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}^2\mathbf{e} = \mathbf{A}(\mathbf{A}\mathbf{e}) = \lambda\mathbf{A}\mathbf{e} = \lambda^2\mathbf{e}\) | M1 A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of \(\mathbf{A}\) are \(n, 2n\) and \(3n\) | B1 | |
| \(\lambda = n\): \(\mathbf{e}_1 = t\begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda = 2n\): \(\mathbf{e}_2 = t\begin{pmatrix}1\\n\\0\end{pmatrix}\) | A1 | |
| \(\lambda = 3n\): \(\mathbf{e}_3 = t\begin{pmatrix}3\\0\\2n\end{pmatrix}\) | A1 | |
| Eigenvalues of \(\mathbf{A} + n\mathbf{I}\) are \(2n, 3n\) and \(4n\) | B1 | |
| \(\mathbf{P} = \begin{pmatrix}1&1&3\\0&n&0\\0&0&2n\end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix}(2n)^2&0&0\\0&(3n)^2&0\\0&0&(4n)^2\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^2\mathbf{e} = \mathbf{A}(\mathbf{A}\mathbf{e}) = \lambda\mathbf{A}\mathbf{e} = \lambda^2\mathbf{e}$ | M1 A1 | AG |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of $\mathbf{A}$ are $n, 2n$ and $3n$ | B1 | |
| $\lambda = n$: $\mathbf{e}_1 = t\begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda = 2n$: $\mathbf{e}_2 = t\begin{pmatrix}1\\n\\0\end{pmatrix}$ | A1 | |
| $\lambda = 3n$: $\mathbf{e}_3 = t\begin{pmatrix}3\\0\\2n\end{pmatrix}$ | A1 | |
| Eigenvalues of $\mathbf{A} + n\mathbf{I}$ are $2n, 3n$ and $4n$ | B1 | |
| $\mathbf{P} = \begin{pmatrix}1&1&3\\0&n&0\\0&0&2n\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}(2n)^2&0&0\\0&(3n)^2&0\\0&0&(4n)^2\end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns |9 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 $\mathbf { A } ^ { 2 }$, with corresponding eigenvalue $\lambda ^ { 2 }$.\\
The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by
$$\mathbf { A } = \left( \begin{array} { c c c }
n & 1 & 3 \\
0 & 2 n & 0 \\
0 & 0 & 3 n
\end{array} \right) \quad \text { and } \quad \mathbf { B } = ( \mathbf { A } + n \mathbf { I } ) ^ { 2 }$$
where $\mathbf { I }$ is the $3 \times 3$ identity matrix and $n$ is a non-zero integer.\\
(ii) Find, in terms of $n$, a non-singular matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { B } = \mathbf { P D P } \mathbf { P } ^ { - 1 }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q9 [10]}}