| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Prove eigenvalue/eigenvector properties |
| Difficulty | Standard +0.8 This is a two-part Further Maths question combining proof and computation. Part (i) is straightforward proof by contradiction, part (ii) requires manipulating the eigenvalue equation (both standard FP1 material). The computational part requires finding eigenvalues of an upper triangular matrix (easy), applying the proven result to find eigenvalues of B, finding eigenvectors, and constructing the diagonalization. While multi-step, each component uses standard techniques taught in FP1, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 4.03a Matrix language: terminology and notation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
The square matrix $\mathbf { A }$ has $\lambda$ as an eigenvalue with $\mathbf { e }$ as a corresponding eigenvector. Show that if $\mathbf { A }$ is non-singular then\\
(i) $\lambda \neq 0$,\\
(ii) the matrix $\mathbf { A } ^ { - 1 }$ has $\lambda ^ { - 1 }$ as an eigenvalue with $\mathbf { e }$ as a corresponding eigenvector.
The $3 \times 3$ matrices $\mathbf { A }$ and $\mathbf { B }$ are given by
$$\mathbf { A } = \left( \begin{array} { r r r }
- 2 & 2 & - 4 \\
0 & - 1 & 5 \\
0 & 0 & 3
\end{array} \right) \quad \text { and } \quad \mathbf { B } = ( \mathbf { A } + 3 \mathbf { I } ) ^ { - 1 }$$
where $\mathbf { I }$ is the $3 \times 3$ identity matrix. Find 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 2014 Q11 OR}}