Hint
An eigenvector is nonzero. Compute \(\displaystyle Q^{−1}AQ\) or apply A to the first column of Q.
Worked solution
(a)
\(Q^{-1}AQ=\left[\begin{matrix}9 & 0 & 0\\0 & 11 & 0\\0 & 0 & 14\end{matrix}\right]\)
Answer: \(9\).
Answer: \(9\)
(b)
\(\prod\lambda_i=1386\)
Answer: \(1386\).
Answer: \(1386\)
(c)
\(\displaystyle A^3=Q D^3Q^{-1}=\left[\begin{matrix}1030 & 301 & -301\\- \frac{1413}{2} & \frac{4075}{2} & \frac{1413}{2}\\- \frac{2015}{2} & \frac{2015}{2} & \frac{3473}{2}\end{matrix}\right]\)
Answer: \(1030\).
Answer: \(1030\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Compute \(\displaystyle Q^{−1}AQ\) or apply A to the first column of Q.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Eigenvectors are columns, not rows, in A=QDQ⁻¹.
(b) · 2 marks
- 1 mark: Multiply the three eigenvalues.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not raise each entry of A separately to compute \(\displaystyle A^{n}\).
(c) · 2 marks
- 1 mark: Cube the diagonal eigenvalues and transform back.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Eigenvectors are columns, not rows, in A=QDQ⁻¹.
Total: 6 marks.