Question

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

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

Answer 1

The correct option is C $\left[\begin{array}{cc}a-ib& -c-id\\ c-id& a+ib\end{array}\right]$

$A=\left[\begin{array}{cc}a+ib& c+id\\ -c+id& a-ib\end{array}\right]$ and $a^2+b^2+c^2+d^2=1$
$|A|=(a+ib)(a-ib)-(-c+id)(c+id)$
$=a^2+b^2+c^2+d^2=1$
Thus, $A^{-1}=\frac{adjA}{|A|}=\left[\begin{array}{cc}a-ib& -c-id\\ c-id& a+ib\end{array}\right]$
Hence, option C.