Correct Answer - Option 2 :
\( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
Concept:
For a 2 x 2 matrix there is a short-cut formula for inverse as given below
\(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\)
Calculation:
A we know that inverse of matrix \(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\)
⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{(2+3i)(2-3i)-1}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{4+9-1}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)
Hence, option 2 is correct.
