Question

If $\alpha +i\beta =\left[\begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array}\right]$,

Find$(\alpha +i\beta {)}^{-1}$

• A. $(\alpha +i\beta {)}^{-1}={\left[\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right]}^{-1}$
• B. $(\alpha +i\beta {)}^{-1}={\left[\begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array}\right]}^{-1}$
• C. $(\alpha +i\beta {)}^{-1}={\left[\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right]}^{-1}$
• D. $(\alpha +i\beta {)}^{-1}={\left[\begin{array}{cc}\alpha & -\beta \\ -\beta & -\alpha \end{array}\right]}^{-1}$

Answer 1

The correct option is B $(\alpha +i\beta {)}^{-1}={\left[\begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array}\right]}^{-1}$

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

$\therefore A^{-1}={\left[\begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array}\right]}^{-1}$

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

Note: We can apply matrix operation for imaginary number in same way as real number.