Question

If $A=\left|\begin{array}{cc}\frac{1}{2}(e^{\alpha }+e^{\alpha })& \frac{1}{2}\left(e^{\alpha }{e}^{\alpha }\right)\\ \frac{1}{2}\left(e^{\alpha }{e}^{\alpha }\right)& \frac{1}{2}(e^{\alpha }+e^{\alpha })\end{array}\right|$ then $A^1$ exists

• A. For all real $\alpha$
• B. For positive real $\alpha$ only
• C. For negative real $\alpha$ only
• D. None of these

Answer 1

Answer: (A)

For all real $\alpha$

$A=\left|\begin{array}{cc}\frac{1}{2}(e^{\alpha }+e^{\alpha })& \frac{1}{2}\left(e^{\alpha }{e}^{\alpha }\right)\\ \frac{1}{2}\left(e^{\alpha }{e}^{\alpha }\right)& \frac{1}{2}(e^{\alpha }+e^{\alpha })\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\left|\begin{array}{cc}(e^{\alpha }+e^{\alpha })& \left(e^{\alpha }{e}^{\alpha }\right)\\ \left(e^{\alpha }{e}^{\alpha }\right)& (e^{\alpha }+e^{\alpha })\end{array}\right|$

$\Rightarrow A=(e^{\alpha }+e^{\alpha }{)}^2-(e^{\alpha }{e}^{\alpha }{)}^2$

$\Rightarrow A=4e^{2\alpha }-e^{4\alpha }$

So, $A$ is correct option.