Question

If A = [α β
γ -α] is such that A^2=I then
(A) 1+α^2+βγ=0
(B) 1-α^2+βγ=0
(C) 1-α^2-βγ=0
(D)1+α^2-βγ=0

Answer 1

Answer:

Watch ing............. ''..........

Answer 2

Given :

∞$\text{If}\:\: A = \left[\begin{array}{cc}\alpha &\beta \\\gamma&-\alpha\end{array}\right] \text{is such that }A^2=I \: \text{then}$

$(A) 1+\alpha^2+\beta\gamma=0\:\:\:\:(B)1-\alpha^2+\beta\gamma=0$

$(C)1-\alpha^2-\beta\gamma=0\:\:\:\:(D)1+\alpha^2-\beta\gamma=0$

Solution :

Given A = $\left[\begin{array}{cc}\alpha &\beta \\\gamma&-\alpha\end{array}\right]$

Such that $A^2=I$

$\implies AA=I$

$\left[\begin{array}{cc}\alpha &\beta \\\gamma&-\alpha\end{array}\right]\left[\begin{array}{cc}\alpha &\beta \\\gamma&-\alpha\end{array}\right] = \left[\begin{array}{cc}1 &0 \\0&1\end{array}\right]$

$\left[\begin{array}{cc}\alpha(\alpha)+\beta(\gamma) &\alpha(\beta)+\beta(-\alpha) \\\gamma(\alpha)-\alpha(\gamma)&\gamma(\beta)-\alpha(-\alpha)\end{array}\right] = \left[\begin{array}{cc}1 &0 \\0&1\end{array}\right]$

$\left[\begin{array}{cc}\alpha^2+\beta\gamma&\alpha\beta -\alpha\beta\\\gamma\alpha-\alpha\gamma&\gamma\beta+\alpha^2\end{array}\right] = \left[\begin{array}{cc}1 &0 \\0&1\end{array}\right]$

$\left[\begin{array}{cc}\alpha^2+\beta\gamma&0 \\0+\beta\gamma&\infty^2\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Equating the like terms

$\implies \alpha^2+\beta\gamma=1$

$\implies 1-\alpha^2-\alpha^2\beta\Gamma=0$

$\implies 1-\alpha^2-\beta\gamma=0$

Final Answer :

$\boxed {Ans:C}$