Question Number 58480 by rahul 19 last updated on 23/Apr/19
Commented by kaivan.ahmadi last updated on 23/Apr/19
![Let P= [((a b)),((c d)) ] [1 0] [((a b)),((c d)) ]=[a b]⇒a=b=((−1)/( (√2))) and [0 1] [((a b)),((c d)) ]=[c d]⇒c=((−1)/( (√2))) , d=(1/( (√2))) ⇒P=((−1)/( (√2))) [((1 1)),((1 −1)) ] P^2 =(1/2) [((1 1)),((1 −1)) ] [((1 1)),((1 −1)) ]=(1/2) [((2 0)),((0 2)) ]=I ⇒P^4 =P^6 =P^8 =I so (C) is true, since P^8 +P^6 +P^4 −P^2 =I+I+I−I=2I](https://www.tinkutara.com/question/Q58494.png)
$${Let}\:{P}=\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix} \\ $$$$\left[\mathrm{1}\:\:\mathrm{0}\right]\begin{bmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{a}\:\:{b}\right]\Rightarrow{a}={b}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${and} \\ $$$$\left[\mathrm{0}\:\:\mathrm{1}\right]\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{c}\:\:\:{d}\right]\Rightarrow{c}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:,\:{d}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{P}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$$${P}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:−\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:−\mathrm{1}}\end{bmatrix}=\frac{\mathrm{1}}{\mathrm{2}}\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{2}}\end{bmatrix}={I} \\ $$$$\Rightarrow{P}^{\mathrm{4}} ={P}^{\mathrm{6}} ={P}^{\mathrm{8}} ={I} \\ $$$${so}\:\left({C}\right)\:{is}\:{true},\:{since} \\ $$$${P}^{\mathrm{8}} +{P}^{\mathrm{6}} +{P}^{\mathrm{4}} −{P}^{\mathrm{2}} ={I}+{I}+{I}−{I}=\mathrm{2}{I} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 24/Apr/19
$${thank}\:{u}\:{sir}. \\ $$