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Mathematics Question on Properties of Determinants

Let P be a 2×22 \times 2 matrix such that [1 0] P = −12[11] - \frac{1}{\sqrt{2}} [ 1 1] and [01]P=12[−11][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

P8−P6+P4+P2=0P^8 - P^6 + P^4 + P^2 = 0

B

P8−P6−P4+P2=IP^8 - P^6 - P^4 + P^2 = I

C

P8+P6+P4−P2=2IP^8 + P^6 + P^4 - P^2 = 2I

D

P8−P6−P4−P2=0P^8 - P^6 - P^4 - P^2 = 0

Answer

P8+P6+P4−P2=2IP^8 + P^6 + P^4 - P^2 = 2I

Explanation

Solution

Let P=[ab cd]P =\begin{bmatrix}a&b\\\ c&d\end{bmatrix}
Given,
∴[10]P=−12[11]\therefore\left[1 0\right]P = \frac{-1}{\sqrt{2}} \left[1 1\right]
[10][ab cd]=[−12−12]\left[10\right]\begin{bmatrix}a&b\\\ c&d\end{bmatrix} = \left[-\frac{1}{\sqrt{2}} \frac{-1}{\sqrt{2}}\right]
[ab]=[−12−12]\left[ab\right] = \left[\frac{-1}{\sqrt{2}} \frac{-1}{\sqrt{2}}\right]
and [01]P=12[−11]\left[ 01\right]P = \frac{1}{\sqrt{2}}\left[-1 1\right]
[01][ab cd]=[−1212]\left[0 1\right]\begin{bmatrix}a&b\\\ c&d\end{bmatrix} = \left[ - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right]
[cd]=[−1212]\left[cd\right] = \left[ -\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right]
∴a=−12,b=−12,c=−12,d=12\therefore a = \frac{-1}{\sqrt{2}}, b= \frac{-1}{\sqrt{2}}, c = \frac{-1}{\sqrt{2}}, d = \frac{1}{\sqrt{2}}
P=12[−1−1 −11]P = \frac{1}{\sqrt{2}}\begin{bmatrix}-1&-1\\\ -1&1\end{bmatrix}
P2=12[−1−1 −11][−1−1 −11]=12[20 02]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}
P2=IP^{2} = I
P4=I=P6=P8P^{4} = I = P^{6} = P^{8}
P8+P6+P4−P4−P2P^{8} +P^{6} +P^{4} -P^{4} -P^{2}
=I+I+I−I=2I= I + I+ I -I = 2I