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

Answer: (B)

$1 - \alpha^2 - \beta{y} = 0$

Answer 2

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
$\alpha^2 + \beta \gamma = 1$
$\Rightarrow 1 - \alpha^2 - \beta \gamma = 0$