If $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,\,(i=\sqrt{-1})$ , then ${{A}^{-1}}=$.
A. $\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$
B. $\left( \begin{matrix}
   -i & 0 \\
   0 & -2i \\
\end{matrix} \right)\,$
C. $\left( \begin{matrix}
   i & 0 \\
   0 & 2i \\
\end{matrix} \right)\,$
D. $\left( \begin{matrix}
   0 & i \\
   2i & 0 \\
\end{matrix} \right)\,$

Hint: In order to determine the value of ${{A}^{-1}}$, first we will calculate the determinant and the adjoint of the matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$. Then we will substitute the values in the formula of calculating the inverse of the matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$. The adjoint of the matrix is calculated by interchanging the element of the principal diagonal and only change the sign of the other diagonal.

Formula Used: \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]
If $A=\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)\,$ then $|A|=ad-bc$.

Complete step by step solution: We are given a matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$and we have to find its inverse ${{A}^{-1}}$ when $i=\sqrt{-1}$.
To find the inverse of the matrix we will use the formula \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\], so we will calculate determinant and adjoint of the matrix \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\] and then substitute in the formula. The value of the determinant must be non-zero because if the determinant is zero then inverse of that matrix will not exist.
First we will find the determinant of the matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$. If the value of the determinant of the matrix
$|A|=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$
$=i\times \dfrac{i}{2}-0\times 0$
$=\dfrac{{{i}^{2}}}{2}$
Substituting the value $i=\sqrt{-1}$.
$=\dfrac{-1}{2}$
We will now determine the adjoint of the matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$.
Interchanging the diagonal of the principal diagonal and changing the sign of the other diagonal,
$adj(A)=\left( \begin{matrix}
   i/2 & -0 \\
   -0 & i \\
\end{matrix} \right)\,$
As Sign before $0$holds no value then,
$adj(A)=\left( \begin{matrix}
   i/2 & 0 \\
   0 & i \\
\end{matrix} \right)\,$
We will determine the inverse by substituting the values.
\[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]
\[{{A}^{-1}}=\dfrac{1}{-1/2}\,\left( \begin{matrix}
   i/2 & 0 \\
   0 & i \\
\end{matrix} \right)\]
We will multiply the matrix by $\dfrac{-1}{2}$,
\[{{A}^{-1}}=\,\left( \begin{matrix}
   -i & 0 \\
   0 & -2i \\
\end{matrix} \right)\]
The inverse of the matrix $A=\left( \begin{matrix}
   i & 0 \\
   0 & i/2 \\
\end{matrix} \right)\,$is \[{{A}^{-1}}=\,\left( \begin{matrix}
   -i & 0 \\
   0 & -2i \\
\end{matrix} \right)\], hence the correct option is (B).

Option ‘B’ is correct

Note: The adjoint of a $2\times 2$ matrix can be directly find by using this method. If there is a matrix $A=\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)\,$then adjoint can be calculated by using this \[adj(A)=\left( \begin{matrix}
   d & -b \\
   -c & a \\
\end{matrix} \right)\].