Hint

An eigenvector is nonzero. Compute Q⁻¹AQ or apply A to the first column of Q.

Worked solution

(a)

\(Q^{-1}AQ=\left[\begin{matrix}10 & 0 & 0\\0 & 12 & 0\\0 & 0 & 15\end{matrix}\right]\)

Answer: \(10\).

Answer: \(10\)

(b)

\(\prod\lambda_i=1800\)

Answer: \(1800\).

Answer: \(1800\)

(c)

\(A^3=Q D^3Q^{-1}=\left[\begin{matrix}1364 & 364 & -364\\- \frac{1647}{2} & \frac{5103}{2} & \frac{1647}{2}\\- \frac{2375}{2} & \frac{2375}{2} & \frac{4375}{2}\end{matrix}\right]\)

Answer: \(1364\).

Answer: \(1364\)

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 Q⁻¹AQ 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 Aⁿ.

(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.