If P = [ * 20 c 3 2 & 1 2 - 1 2 & 3 2 ], ,A = , [ * 20 c 1&1 0&1 … - Vidyadip

MCQ

If $P = \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right],\,A = \,\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $Q=PAP^T,$ then $P^T$ $Q^{2015}$ $P$ is

A
$\,\left[ {\begin{array}{*{20}{c}}
0&{2015}\\
0&0
\end{array}} \right]$
B
$\,\left[ {\begin{array}{*{20}{c}}
{2015}&0\\
1&{2015}
\end{array}} \right]$
✓
$\left[ {\begin{array}{*{20}{c}}
1&{2015}\\
0&1
\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}
{2015}&1\\
0&{2015}
\end{array}} \right]$

✓

Answer

Correct option: C.

$\left[ {\begin{array}{*{20}{c}}
1&{2015}\\
0&1
\end{array}} \right]$

c
$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

$P^{T}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

$P P^{T}=P^{T} P=I$

$\mathrm{Q}^{2015}=\left(P A P^{T}\right)\left(P A P^{T}\right)-(2015 \text { terms })$

$=P A^{2015} P^{T}$

$P^{T} \mathrm{Q}^{2015} P=A^{2015}$

$A^{2}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$

$A^{3}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$

$A^{2015}=\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]$

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