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Question-58480




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
$${Let}\:{P}=\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix} \\ $$$$\left[\mathrm{1}\:\:\mathrm{0}\right]\begin{bmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{a}\:\:{b}\right]\Rightarrow{a}={b}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${and} \\ $$$$\left[\mathrm{0}\:\:\mathrm{1}\right]\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{c}\:\:\:{d}\right]\Rightarrow{c}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:,\:{d}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{P}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$$${P}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:−\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:−\mathrm{1}}\end{bmatrix}=\frac{\mathrm{1}}{\mathrm{2}}\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{2}}\end{bmatrix}={I} \\ $$$$\Rightarrow{P}^{\mathrm{4}} ={P}^{\mathrm{6}} ={P}^{\mathrm{8}} ={I} \\ $$$${so}\:\left({C}\right)\:{is}\:{true},\:{since} \\ $$$${P}^{\mathrm{8}} +{P}^{\mathrm{6}} +{P}^{\mathrm{4}} −{P}^{\mathrm{2}} ={I}+{I}+{I}−{I}=\mathrm{2}{I} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 24/Apr/19
thank u sir.
$${thank}\:{u}\:{sir}. \\ $$

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