COMEDK · Maths · 21. Matrices
\(\text { If } A=\left(\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right) \quad P=\left(\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right) \quad Q=P^T A P, \quad \text { then } P Q^{2014} P^T \text { is equal to }\)
- A \(\left(P^T\right)^{2013} A^{2014} P^{2013}\)
- B \(\left(\begin{array}{cc}
1 & 2^{2014} \\
0 & 1
\end{array}\right)\) - C \(P^T A^{2014} P\)
- D \(\left(\begin{array}{cc}
1 & 4028 \\
0 & 1
\end{array}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\begin{array}{cc}
1 & 4028 \\
0 & 1
\end{array}\right)\)
Step-by-step Solution
Detailed explanation
Given \(Q = P^T A P\). We need to evaluate \(P Q^{2014} P^T\). Note that \(Q^2 = (P^T A P)(P^T A P) = P^T A (P P^T) A P\). Since \(P\) is an orthogonal matrix, \(P P^T = I\), where \(I\) is the identity matrix. Thus, \(Q^2 = P^T A^2 P\). By induction, \(Q^n = P^T A^n P\) for any…
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