First slide

Transpose of matrix

Question

Let $P = [\begin{array}{cc} \sqrt{3} / 2 & 1 / 2 \\ â 1 / 2 & \sqrt{3} / 2 \end{array}], A = [\begin{array}{ll} 1 ââââ 1 \\ 0 ââââ 1 \end{array}]$ and $Q = P A P^{â²}$ then $P^{â²} Q^{2015} P$ is

Moderate

A

$[\begin{array}{cc} 1 & â 2015 \\ 0 & 1 \end{array}]$
B

$[\begin{array}{cc} 2015 & 1 \\ 0 & 2015 \end{array}]$
C

$[\begin{array}{cc} 1 & 2015 \\ 0 & 1 \end{array}]$
D

$[\begin{array}{cc} 2015 & 2015 \\ 0 & 2015 \end{array}]$

Solution

$P = [\begin{array}{cc} \cos â¡ (Ï / 6) & \sin â¡ (Ï / 6) \\ â \sin â¡ (Ï / 6) & \cos â¡ (Ï / 6) \end{array}]$
$â P^{â²} = [\begin{array}{cc} \cos â¡ (Ï / 6) & â \sin â¡ (Ï / 6) \\ \sin â¡ (Ï / 6) & \cos â¡ (Ï / 6) \end{array}]$
Since $P P^{â²} = [\begin{array}{ll} 1 ââââ 0 \\ 0 ââââ 1 \end{array}] â P^{â²} = P^{â 1}$
We have $Q = P A P^{â²} = P A P^{â 1}$
$â Q^{2015} = {(P A P^{â 1})}^{2015} = P A^{2015} P^{â 1}$
Thus, $P^{â²} Q^{2015} P = P^{â 1} (P A^{2015} P^{â 1}) P$
$= (P^{â 1} P) A^{2015} (P^{â 1} P)$
Now, $A = I + B$ where $B = [\begin{array}{ll} 0 ââââ 1 \\ 0 ââââ 0 \end{array}]$
Since, $B^{2} = O$ , we get $B^{r} = O â r â¥ 2$
Thus, $A^{2015} = I + 2015 B = [\begin{array}{cc} 1 & 2015 \\ 0 & 1 \end{array}]$

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