Matrices & Determinants
Matrices
nta_pyq_2025_jan
Grade 12
Question:
Let $A=\begin{pmatrix}-\dfrac{1}{\sqrt{2}} & 1\\ 0 & 1\end{pmatrix}$ and $P=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix},\,\theta>0.$ If $B=PAP^{T},\,C=P^{T}B^{10}P$ and the sum of the diagonal elements of $C$ is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is:
Step-by-Step Solution
Key Concept: $P$ is orthogonal: $P^{T}P=I.$ So $B=PAP^{T}\Rightarrow B^{10}=PA^{10}P^{T}\Rightarrow C=P^{T}B^{10}P=A^{10}.$ Then trace$(C)=$ trace$(A^{10})=\sum\lambda_{i}^{10}.$
$B^{10}=(PAP^{T})^{10}=PA^{10}P^{T}$ (since $P^{T}P=I$).
$C=P^{T}B^{10}P=P^{T}PA^{10}P^{T}P=A^{10}.$
$A$ is upper triangular with eigenvalues $-\dfrac{1}{\sqrt{2}}$ and $1.$ Trace$(A^{10})=\left(-\dfrac{1}{\sqrt{2}}\right)^{10}+1^{10}=\dfrac{1}{32}+1=\dfrac{33}{32}.$
$\gcd(33,32)=1\Rightarrow m+n=65.$
Correct Answer: 3