Question

For a given matrix $P=\left[\begin{array}{cc}4+3i& -i\\ i& 4-3i\end{array}\right]$, $i=\sqrt{-}1$, the inverse of matrix P is

• A. $\frac{1}{24}$$\left[\begin{array}{cc}4-3i& i\\ -i& 4+3i\end{array}\right]$
• B. $\frac{1}{25}$$\left[\begin{array}{cc}i& 4-3i\\ 4+3i& -i\end{array}\right]$
• C. $\frac{1}{24}$$\left[\begin{array}{cc}4+3i& -i\\ i& 4-3i\end{array}\right]$
• D. $\frac{1}{25}$$\left[\begin{array}{cc}4+3i& -i\\ i& 4-3i\end{array}\right]$
  

Answer 1

The correct option is A $\frac{1}{24}$$\left[\begin{array}{cc}4-3i& i\\ -i& 4+3i\end{array}\right]$

$P=\left[\begin{array}{cc}4+3i& -i\\ i& 4-3i\end{array}\right]$, i = $\sqrt{-}1$

We know that,
${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{(ad-bc)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
So,

${\left[\begin{array}{cc}4+3i& -i\\ i& 4-3i\end{array}\right]}^{-1}=\frac{1}{\left[\right(4+3i\left)\right(4-3i+i^2\left)\right]}\times \left[\begin{array}{cc}4-3i& i\\ -i& 4+3i\end{array}\right]$
=$\frac{1}{[16-9i^2+i^2]}$$\left[\begin{array}{cc}4-3i& i\\ -i& 4+3i\end{array}\right]$
$=\frac{1}{24}$$\left[\begin{array}{cc}4-3i& i\\ -i& 4+3i\end{array}\right]$

Option (a) is correct.