Question

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

• A. for all real $x$
• B. for positive real $x$ only
• C. for negative real $x$ only
• D. none of these

Answer 1

The correct option is A for all real $x$

$A=\left[\begin{array}{cc}\frac{1}{2}(e^{ix}+e^{-ix})& \frac{1}{2}(e^{ix}-e^{-ix})\\ \frac{1}{2}(e^{ix}-e^{-ix})& \frac{1}{2}(e^{ix}+e^{-ix})\end{array}\right]$

$\Rightarrow A=\left[\begin{array}{cc}coshx& sinhx\\ sinhx& coshx\end{array}\right]$

$|A|=cos{h}^{2}x-sin{h}^{2}x=1$

$\therefore A^{-1}$ exists or all $x$

Hence, option A.