CAIE FP1 2012 November — Question 10 13 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A = PDP⁻¹
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring eigenvalue/eigenvector calculation for a triangular matrix (straightforward), diagonalization, matrix inversion, and a limit involving matrix powers. While the triangular structure simplifies eigenvalue identification, the full diagonalization process, finding P⁻¹, computing Aⁿ, and the limit calculation require solid technique across multiple steps. This is moderately challenging for Further Maths FP1 level, above average difficulty overall.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)

10 Write down the eigenvalues of the matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { r r r } 1 & 4 & - 16 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{array} \right)$$ Find corresponding eigenvectors. Let \(n\) be a positive integer. Write down a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that $$\mathbf { A } ^ { n } = \mathbf { P D } \mathbf { P } ^ { - 1 }$$ Find \(\mathbf { P } ^ { - 1 }\) and \(\mathbf { A } ^ { n }\). Hence find \(\lim _ { n \rightarrow \infty } \left( 3 ^ { - n } \mathbf { A } ^ { n } \right)\).

B1 States eigenvalues are 1, 2, 3.
M1A1 Finds eigenvector for \(\lambda = 1\): \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)
A1 Finds eigenvector for \(\lambda = 2\): \(\mathbf{e}_2 = \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}\)
A1 Finds eigenvector for \(\lambda = 3\): \(\mathbf{e}_3 = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)
B1 States \(P = \begin{pmatrix} 1 & 4 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}\)
B1 States \(D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix}\)
M1A1 Finds \(\det P = 1\) and \(P^{-1} = \begin{pmatrix} 1 & -4 & 14 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}\)
M1A1 Finds \(A^n = PD^nP^{-1} = \begin{pmatrix} 1 & 4 \cdot 2^n - 2 \cdot 3^n & -3 \cdot 2^n + 14 \cdot 3^n \\ 0 & 2^n & -3 \cdot 2^n + 3^{n+1} \\ 0 & 0 & 3^n \end{pmatrix}\)
A1 States \(3^{-n}A^n \to \begin{pmatrix} 0 & 0 & -2 \\ 0 & 0 & 3 \\ 0 & 0 & 1 \end{pmatrix}\) as \(n \to \infty\)
B1 States required limit.
B1 States eigenvalues are 1, 2, 3.

M1A1 Finds eigenvector for $\lambda = 1$: $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

A1 Finds eigenvector for $\lambda = 2$: $\mathbf{e}_2 = \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}$

A1 Finds eigenvector for $\lambda = 3$: $\mathbf{e}_3 = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}$

B1 States $P = \begin{pmatrix} 1 & 4 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$

B1 States $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix}$

M1A1 Finds $\det P = 1$ and $P^{-1} = \begin{pmatrix} 1 & -4 & 14 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$

M1A1 Finds $A^n = PD^nP^{-1} = \begin{pmatrix} 1 & 4 \cdot 2^n - 2 \cdot 3^n & -3 \cdot 2^n + 14 \cdot 3^n \\ 0 & 2^n & -3 \cdot 2^n + 3^{n+1} \\ 0 & 0 & 3^n \end{pmatrix}$

A1 States $3^{-n}A^n \to \begin{pmatrix} 0 & 0 & -2 \\ 0 & 0 & 3 \\ 0 & 0 & 1 \end{pmatrix}$ as $n \to \infty$

B1 States required limit.
10 Write down the eigenvalues of the matrix $\mathbf { A }$, where

$$\mathbf { A } = \left( \begin{array} { r r r } 
1 & 4 & - 16 \\
0 & 2 & 3 \\
0 & 0 & 3
\end{array} \right)$$

Find corresponding eigenvectors.

Let $n$ be a positive integer. Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that

$$\mathbf { A } ^ { n } = \mathbf { P D } \mathbf { P } ^ { - 1 }$$

Find $\mathbf { P } ^ { - 1 }$ and $\mathbf { A } ^ { n }$.

Hence find $\lim _ { n \rightarrow \infty } \left( 3 ^ { - n } \mathbf { A } ^ { n } \right)$.

\hfill \mbox{\textit{CAIE FP1 2012 Q10 [13]}}
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