Question

Let P be a $2 \times 2$ matrix such that [1 0] P = $ - \frac{1}{\sqrt{2}} [ 1 1]$ and $[0 1] P = \frac{1}{\sqrt{2}} [- 1 1]$ . 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

The Correct Option is C

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