| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| Paper | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Prove eigenvalue/eigenvector properties |
| Difficulty | Challenging +1.2 This question combines standard eigenvalue theory proofs with routine matrix diagonalization. Part (i) is a straightforward proof by contradiction using det(A)≠0, and part (ii) follows directly from the eigenvalue equation. The computational part involves finding eigenvalues of a triangular matrix (trivial), finding eigenvectors (standard), and writing the diagonalization. While it requires multiple techniques and careful algebraic manipulation, all steps are textbook procedures with no novel insight required. The proof elements are accessible and the matrix work is routine for Further Maths students. |
| Spec | 4.03a Matrix language: terminology and notation4.03o Inverse 3x3 matrix |
The 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 matrices $\mathbf { A }$ and $\mathbf { B }$ are given by
$$\mathbf { A } = \left( \begin{array} { r r r }
1 & - 1 & 2 \\
0 & - 2 & 4 \\
0 & 0 & - 3
\end{array} \right) \quad \text { and } \quad \mathbf { B } = ( \mathbf { A } + 4 \mathbf { I } ) ^ { - 1 }$$
Find a non-singular matrix $\mathbf { P }$, and a diagonal matrix $\mathbf { D }$, such that $\mathbf { B } = \mathbf { P D P } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE FP1 2004 Q12 OR}}