Question

Let P be a $2\times 2$ matrix such that [1 0] P = $-\frac{1}{\sqrt{2}}[11]$ and $[01]P=\frac{1}{\sqrt{2}}[-11]$ . If 0 and I denote the zero and identity matrices of order 2, respectively, then which of the following options is CORRECT ?

• A. $P^8-P^6+P^4+P^2=0$
• B. $P^8-P^6-P^4+P^2=I$
• C. $P^8+P^6+P^4-P^2=2I$
• D. $P^8-P^6-P^4-P^2=0$

Answer 1

Answer: (C)

Let $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Given,
$\therefore \left[10\right]P=\frac{-1}{\sqrt{2}}\left[11\right]$
$\left[10\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[-\frac{1}{\sqrt{2}}\frac{-1}{\sqrt{2}}\right]$
$\left[ab\right]=\left[\frac{-1}{\sqrt{2}}\frac{-1}{\sqrt{2}}\right]$
and $\left[01\right]P=\frac{1}{\sqrt{2}}\left[-11\right]$
$\left[01\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\right]$
$\left[cd\right]=\left[-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\right]$
$\therefore a=\frac{-1}{\sqrt{2}},b=\frac{-1}{\sqrt{2}},c=\frac{-1}{\sqrt{2}},d=\frac{1}{\sqrt{2}}$
$P=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right]$
$P^2=\frac{1}{2}\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$
$P^2=I$
$P^4=I=P^6=P^8$
$P^8+P^6+P^4-P^4-P^2$
$=I+I+I-I=2I$