In [1, p. 35 exercise 2.18] we are asked to prove that the inverse of a complex matrix \mathbf{A}=\mathbf{A}_{R}+j\mathbf{A}_{I} may be found by first inverting
\mathbf{V}=\left( \begin{array}{cc} \mathbf{A}_{R} & -\mathbf{A}_{I} \\ \mathbf{A}_{I} & \mathbf{A}_{R} \\ \end{array} \right) (1)

to yield
\mathbf{V}^{-1}=\left( \begin{array}{cc} \mathbf{B}_{R} & -\mathbf{B}_{I} \\ \mathbf{B}_{I} & \mathbf{B}_{R} \\ \end{array} \right) (2)

and then letting \mathbf{A}^{-1}=\mathbf{B}_{R}+j\mathbf{B}_{I}.
Solution: From the information that (2) is the inverse of (1) we can obtain the relations:
\mathbf{V}\mathbf{V}^{-1}=\mathbf{I}
\left( \begin{array}{cc} \mathbf{A}_{R} & -\mathbf{A}_{I} \\ \mathbf{A}_{I} & \mathbf{A}_{R} \\ \end{array} \right) \cdot \left( \begin{array}{cc} \mathbf{B}_{R} & -\mathbf{B}_{I} \\ \mathbf{B}_{I} & \mathbf{B}_{R} \\ \end{array} \right)=\left( \begin{array}{cc} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{array} \right)
\left( \begin{array}{cc} \mathbf{A}_{R}\mathbf{B}_{R} - \mathbf{A}_{I}\mathbf{B}_{I} & -\mathbf{A}_{R}\mathbf{B}_{I} - \mathbf{A}_{I}\mathbf{B}_{R} \\ \mathbf{A}_{R}\mathbf{B}_{I} + \mathbf{A}_{I}\mathbf{B}_{R} & \mathbf{A}_{R}\mathbf{B}_{R} - \mathbf{A}_{I}\mathbf{B}_{I} \\ \end{array} \right)=\left( \begin{array}{cc} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{array} \right)

which can be reduced to :
\mathbf{A}_{R}\mathbf{B}_{R} - \mathbf{A}_{I}\mathbf{B}_{I}=\mathbf{I} (3)
\mathbf{A}_{R}\mathbf{B}_{I} + \mathbf{A}_{I}\mathbf{B}_{R}=\mathbf{0} (4)

Computing the product \mathbf{V}^{-1}\mathbf{V} will result in:
\mathbf{V}^{-1} \mathbf{V}=\mathbf{I}
\left( \begin{array}{cc} \mathbf{B}_{R} & -\mathbf{B}_{I} \\ \mathbf{B}_{I} & \mathbf{B}_{R} \\ \end{array} \right) \cdot \left( \begin{array}{cc} \mathbf{A}_{R} & -\mathbf{A}_{I} \\ \mathbf{A}_{I} & \mathbf{A}_{R} \\ \end{array} \right)=\left( \begin{array}{cc} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{array} \right)
\left( \begin{array}{cc} \mathbf{B}_{R}\mathbf{A}_{R} - \mathbf{B}_{I}\mathbf{A}_{I} & -\mathbf{B}_{R}\mathbf{A}_{I} - \mathbf{B}_{I}\mathbf{A}_{R} \\ \mathbf{B}_{R}\mathbf{A}_{I} + \mathbf{B}_{I}\mathbf{A}_{R} & \mathbf{B}_{R}\mathbf{A}_{R} - \mathbf{B}_{I}\mathbf{A}_{I} \\ \end{array} \right)=\left( \begin{array}{cc} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{array} \right)

and the relations that can be obtained by the previous equation can be reduced to :
\mathbf{B}_{R}\mathbf{A}_{R} - \mathbf{B}_{I}\mathbf{A}_{I}=\mathbf{I} (5)
\mathbf{B}_{R}\mathbf{A}_{I} + \mathbf{B}_{I}\mathbf{A}_{R}=\mathbf{0} (6)

Now we will proceed to show that the inverse of \mathbf{A}=\mathbf{A}_{R}+j\mathbf{A}_{I} is \mathbf{B}=\mathbf{B}_{R}+j\mathbf{B}_{I}. First we will proof that \mathbf{B} is the right inverse of \mathbf{A}:
\mathbf{A}\mathbf{B}=(\mathbf{A}_{R}+j\mathbf{A}_{I}) (\mathbf{B}_{R}+j\mathbf{B}_{I})
=\mathbf{A}_{R}\mathbf{B}_{R}+j\mathbf{A}_{R}\mathbf{B}_{I} +j\mathbf{A}_{I}\mathbf{B}_{R}-\mathbf{A}_{I}\mathbf{B}_{I}
=(\mathbf{A}_{R}\mathbf{B}_{R}-\mathbf{A}_{I}\mathbf{B}_{I})+j(\mathbf{A}_{R}\mathbf{B}_{I} +\mathbf{A}_{I}\mathbf{B}_{R})
=\mathbf{I} +\mathbf{0}

The last relation is obtained by considering (3) and (4). Next we will show that \mathbf{B} is also the left inverse of \mathbf{A}:
\mathbf{B}\mathbf{A}=(\mathbf{B}_{R}+j\mathbf{B}_{I}) (\mathbf{A}_{R}+j\mathbf{A}_{I})
=\mathbf{B}_{R}\mathbf{A}_{R}+j\mathbf{B}_{R}\mathbf{A}_{I} +j\mathbf{B}_{I}\mathbf{A}_{R}-\mathbf{B}_{I}\mathbf{A}_{I}
=(\mathbf{B}_{R}\mathbf{A}_{R}-\mathbf{B}_{I}\mathbf{A}_{I})+j(\mathbf{B}_{R}\mathbf{A}_{I} +\mathbf{B}_{I}\mathbf{A}_{R})
=\mathbf{I} +\mathbf{0}

The last relation is obtained by considering (5) and (6). Thus the matrix \mathbf{B} is the inverse of the matrix \mathbf{A}. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.