Matrices & Determinants
Matrix Diagonalization / Orthogonal Transformation
nta_pyq_2025_apr
Grade 12
Question:
Let $A = \begin{bmatrix}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$, $\theta > 0$. If $B = PAP^T$, $C = P^TB^{10}P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $m + n$ is:
Step-by-Step Solution
Key Concept: Since $P$ is orthogonal ($P^TP = I$), $C = P^TB^{10}P = P^T(PAP^T)^{10}P = A^{10}$. Compute $A^{10}$ directly.
$C = P^T B^{10} P = A^{10}$. Computing: $A^2 = \begin{bmatrix}1/2 & -\sqrt{2}-2 \\ 0 & 1\end{bmatrix}$, and by induction $A^{10}$ has diagonal entries $\left(\frac{1}{\sqrt{2}}\right)^{10} = \frac{1}{32}$ and $1$. Sum $= \frac{1}{32} + 1 = \frac{33}{32} = \frac{m}{n}$. $m + n = 65$.
Correct Answer: 65