If we are given a matrix as $A=\left[ \begin{matrix}
   a+ib & c+id \\
   -c+id & a-ib \\
\end{matrix} \right]$ and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}=1$ , then ${{A}^{-1}}$ is equal to
(A)$\left[ \begin{matrix}
   a+ib & -c-id \\
   -c+id & a-ib \\
\end{matrix} \right]$
(B)$\left[ \begin{matrix}
   a+ib & -c+id \\
   -c+id & a-ib \\
\end{matrix} \right]$
(C)$\left[ \begin{matrix}
   a-ib & -c-id \\
   c-id & a+ib \\
\end{matrix} \right]$
(D) None of these

Hint: For answering this question we will use ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$ and we will initially find the determinant of the matrix using $A=\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ is given by $\left| A \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$ and later we will find the adjoint of the matrix using $adj\left( A \right)=\left[ \begin{matrix}
   {{a}_{22}} & -{{a}_{12}} \\
   -{{a}_{21}} & {{a}_{11}} \\
\end{matrix} \right]$ and further simplify them.

Complete step-by-step solution:
Here in the question it is given that $A=\left[ \begin{matrix}
   a+ib & c+id \\
   -c+id & a-ib \\
\end{matrix} \right]$.
The minor of ${{a}_{ij}}$ is represented by ${{M}_{ij}}$ and for example for any $3\times 3$ matrix the minor of ${{a}_{21}}$ is represented by ${{M}_{21}}$ and is given by $\left| \begin{matrix}
   {{a}_{12}} & {{a}_{13}} \\
   {{a}_{32}} & {{a}_{23}} \\
\end{matrix} \right|$.
The cofactor of ${{a}_{ij}}$ is represented by ${{C}_{ij}}$ is given by ${{C}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}$ .
The cofactor matrix is formed with all the respective cofactors as the elements.
The transpose of a matrix $A$ is given by interchanging all the rows of the matrix with the respective columns.
The adjoint of a matrix $A$ is given as the transpose of the cofactor matrix.
The determinant of a matrix $A=\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ is given by $\left| A \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$ .
By using this for the given matrix we will have $\left| A \right|=\left( a+ib \right)\left( a-ib \right)-\left( c+id \right)\left( -c+id \right)$ .
By simplifying this we will have $\left| A \right|={{a}^{2}}+{{b}^{2}}-\left( -{{c}^{2}}-{{d}^{2}} \right)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}$ .
As it is given in the question that the value of the expression is ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}=1$ using this we will have $\left| A \right|=1$ .
Since we know that the inverse of a matrix is given by ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$ . We need to find the value of adjoint for the sake of finding the inverse.
The adjoint of a matrix $A=\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ is given as the transpose of the cofactor matrix.
The cofactor of ${{a}_{11}}$ is given as ${{a}_{22}}$ .
The cofactor of ${{a}_{21}}$ is given as $-{{a}_{12}}$ .
The cofactor of ${{a}_{12}}$ is given as $-{{a}_{21}}$ .
The cofactor of ${{a}_{22}}$ is given as ${{a}_{11}}$.
Now we have the cofactor matrix as $\left[ \begin{matrix}
   {{a}_{22}} & -{{a}_{21}} \\
   -{{a}_{12}} & {{a}_{11}} \\
\end{matrix} \right]$ .
The transpose of the cofactor matrix that is the adjoint matrix is given as $\left[ \begin{matrix}
   {{a}_{22}} & -{{a}_{12}} \\
   -{{a}_{21}} & {{a}_{11}} \\
\end{matrix} \right]$ .
Hence we will have the adjoint of the given matrix as $adj\left( A \right)=\left[ \begin{matrix}
   a-ib & -c-id \\
   c-id & a+ib \\
\end{matrix} \right]$.
Hence the inverse of the given matrix using $adj\left( A \right)=\left[ \begin{matrix}
   a-ib & -c-id \\
   c-id & a+ib \\
\end{matrix} \right]$ and $\left| A \right|=1$ is given as ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}=\dfrac{\left[ \begin{matrix}
   a-ib & -c-id \\
   c-id & a+ib \\
\end{matrix} \right]}{1}$.
 Hence we end up with a conclusion that the inverse of $A=\left[ \begin{matrix}
   a+ib & c+id \\
   -c+id & a-ib \\
\end{matrix} \right]$ where ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}=1$ will be ${{A}^{-1}}=\left[ \begin{matrix}
   a-ib & -c-id \\
   c-id & a+ib \\
\end{matrix} \right]$.
Hence, option C is correct.

Note: While answering questions of this type we should take care that the while deriving the adjoint of the matrix it is given as $adj\left( A \right)=\left[ \begin{matrix}
   {{a}_{22}} & -{{a}_{12}} \\
   -{{a}_{21}} & {{a}_{11}} \\
\end{matrix} \right]$ not $adj\left( A \right)=\left[ \begin{matrix}
   {{a}_{11}} & -{{a}_{12}} \\
   -{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ if we take this we will get the answer as ${{A}^{-1}}=\left[ \begin{matrix}
   a+ib & -c-id \\
   c-id & a-ib \\
\end{matrix} \right]$. Then we will end up having a wrong answer as option A.