CAIE FP1 2005 November — Question 10 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionNovember
Marks11
PaperDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A = PDP⁻¹
DifficultyStandard +0.8 This is a standard Further Maths eigenvalue/diagonalization question with eigenvalues given, requiring routine eigenvector calculation, matrix diagonalization, and a limit involving powers of matrices. The final limit calculation requires recognizing that only the largest eigenvalue (4) survives when factored out, which is a moderately sophisticated insight beyond pure mechanical computation. The multi-step nature and the conceptual understanding needed for the limit pushes this slightly above average difficulty for Further Maths content.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix

It is given that the eigenvalues of the matrix \(\mathbf{M}\), where $$\mathbf{M} = \begin{pmatrix} 4 & 1 & -1 \\ -4 & -1 & 4 \\ 0 & -1 & 5 \end{pmatrix},$$ are \(1, 3, 4\). Find a set of corresponding eigenvectors. [4] Write down a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that $$\mathbf{M}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1},$$ where \(n\) is a positive integer. [2] Find \(\mathbf{P}^{-1}\) and deduce that $$\lim_{n \to \infty} 4^{-n}\mathbf{M}^n = \begin{pmatrix} -\frac{1}{3} & 0 & -\frac{1}{3} \\ \frac{4}{3} & 0 & \frac{4}{3} \\ \frac{1}{3} & 0 & \frac{1}{3} \end{pmatrix}.$$ [5]

It is given that the eigenvalues of the matrix $\mathbf{M}$, where
$$\mathbf{M} = \begin{pmatrix} 4 & 1 & -1 \\ -4 & -1 & 4 \\ 0 & -1 & 5 \end{pmatrix},$$
are $1, 3, 4$. Find a set of corresponding eigenvectors. [4]

Write down a matrix $\mathbf{P}$ and a diagonal matrix $\mathbf{D}$ such that
$$\mathbf{M}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1},$$
where $n$ is a positive integer. [2]

Find $\mathbf{P}^{-1}$ and deduce that
$$\lim_{n \to \infty} 4^{-n}\mathbf{M}^n = \begin{pmatrix} -\frac{1}{3} & 0 & -\frac{1}{3} \\ \frac{4}{3} & 0 & \frac{4}{3} \\ \frac{1}{3} & 0 & \frac{1}{3} \end{pmatrix}.$$ [5]

\hfill \mbox{\textit{CAIE FP1 2005 Q10 [11]}}