Question

The inverse of the matrix $\left[\begin{array}{cc}3+2i& i\\ -i& 3-2i\end{array}\right]$ is

• A. $\frac{1}{12}$ $\left[\begin{array}{cc}3+2i& -i\\ i& 3-2i\end{array}\right]$
• B. $\frac{1}{12}$ $\left[\begin{array}{cc}3-2i& -i\\ i& 3+2i\end{array}\right]$
• C. $\frac{1}{14}$ $\left[\begin{array}{cc}3+2i& -i\\ i& 3-2i\end{array}\right]$
• D. $\frac{1}{14}$ $\left[\begin{array}{cc}3-2i& -i\\ i& 3+2i\end{array}\right]$

Answer 1

The correct option is B $\frac{1}{12}$ $\left[\begin{array}{cc}3-2i& -i\\ i& 3+2i\end{array}\right]$

Let A = $\left[\begin{array}{cc}3+2i& i\\ -i& 3-2i\end{array}\right]$

$\left|A\right|$ = (3+2i) (3-2i) - (-i) (i)

= (9+4) - (1) = 12

$A^{-1}=\frac{1}{12}\left[\begin{array}{cc}3-2i& -i\\ -(-i)& 3+2i\end{array}\right]$

= $\frac{1}{12}\left[\begin{array}{cc}3-2i& -i\\ +i& 3+2i\end{array}\right]$