Question

Find the inverse of the matrix
$A=\left[\begin{array}{cc}a+ib& c+id\\ -c+id& a-ib\end{array}\right]$, if $a^2+b^2+c^2+d^2=1$

• A. $A^{-1}=\left[\begin{array}{cc}a+ib& -c+id\\ c+id& a-ib\end{array}\right]$
• B. $A^{-1}=\left[\begin{array}{cc}a-ib& -c-id\\ c-id& a+ib\end{array}\right]$
• C. $A^{-1}=\left[\begin{array}{cc}-a-ib& -c+id\\ c-id& a-ib\end{array}\right]$
• D. None of these

Answer 1

The correct option is B $A^{-1}=\left[\begin{array}{cc}a-ib& -c-id\\ c-id& a+ib\end{array}\right]$

$A=\left[\begin{array}{cc}a+ib& c+id\\ -c+id& a-ib\end{array}\right]$ and $a^2+b^2+c^2+d^2=1$

Inverse of $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{cc}a-ib& -c-id\\ c-id& a+ib\end{array}\right]$