Q62 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
$\text { If } A=\left(\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right) \quad P=\left(\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right) \quad Q=P^T A P, \quad \text { then } P Q^{2014} P^T \text { is equal to }$
- A$\left(P^T\right)^{2013} A^{2014} P^{2013}$
- B$\left(\begin{array}{cc} 1 & 2^{2014} \\ 0 & 1 \end{array}\right)$
- C$P^T A^{2014} P$
- Correct answerD$\left(\begin{array}{cc} 1 & 4028 \\ 0 & 1 \end{array}\right)$
Solution
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 positive integer $n$.
Therefore, $Q^{2014} = P^T A^{2014} P$.
Substituting this into the expression $P Q^{2014} P^T$, we get $P (P^T A^{2014} P) P^T$.
Using the associative property of matrix multiplication, $(P P^T) A^{2014} (P P^T) = I A^{2014} I = A^{2014}$.
For the matrix $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, we observe $A^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix}$.
Thus, $A^{2014} = \begin{pmatrix} 1 & 2 \times 2014 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix}$.
Answer: $\left(\begin{array}{cc} 1 & 4028 \\ 0 & 1 \end{array}\right)$
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 positive integer $n$.
Therefore, $Q^{2014} = P^T A^{2014} P$.
Substituting this into the expression $P Q^{2014} P^T$, we get $P (P^T A^{2014} P) P^T$.
Using the associative property of matrix multiplication, $(P P^T) A^{2014} (P P^T) = I A^{2014} I = A^{2014}$.
For the matrix $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, we observe $A^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix}$.
Thus, $A^{2014} = \begin{pmatrix} 1 & 2 \times 2014 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix}$.
Answer: $\left(\begin{array}{cc} 1 & 4028 \\ 0 & 1 \end{array}\right)$
Hence, the correct answer is option: 'D'.