Question

If $ \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix} $ is to be square root of the two rowed unit matrix, then $ \alpha $ , $ \beta $ and $ \gamma $ should statisfy the relation

• A. $ 1 + \alpha^2 + \beta{y} = 0 $
• B. $ 1 - \alpha^2 - \beta{y} = 0 $
• C. $ 1 - \alpha^2 + \beta{y} = 0 $
• D. $ 1 + \alpha^2 - \beta{y} = 0 $

Answer 1

The Correct Option is B

We have,
$\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix} = \begin{bmatrix}1&0\\ 0&1\end{bmatrix}^{1/2} $
$ \Rightarrow \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}^{2} = \begin{bmatrix}1&0\\ 0&1\end{bmatrix}$
[Taking square both sides ]
$\Rightarrow \begin{bmatrix}\alpha^{2}+\beta\gamma&\alpha\beta -\alpha\beta\\ \alpha\gamma-\alpha\gamma&\beta\gamma+\alpha^{2}\end{bmatrix} = \begin{bmatrix}1&0\\ 0&1\end{bmatrix} $ $\Rightarrow\begin{bmatrix}\alpha^{2}+\beta\gamma&0\\ 0&\alpha^{2}+\beta\gamma\end{bmatrix} = \begin{bmatrix}1&0\\ 0&1\end{bmatrix}$
On comparing both sides, we get
$\alpha^2 + \beta \gamma = 1$
$\Rightarrow 1 - \alpha^2 - \beta \gamma = 0$