Prove identities for matrices. Let $A,B,C,D$ be $n \times n$ real valued matrices such that $$AC-BD=Id$$ and $$AD+BC=O$$
Prove $$CA-DB=Id$$ and $$DA+CB=0$$
One idea that I had is to consider the matrices $A,B,C,D$ as blocks of a larger matrix.
Consider the matrix $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} $$ and $$ \begin{pmatrix} C & D \\ D & -C \end{pmatrix} $$
Then $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & D \\ D & -C \end{pmatrix} =\begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$
So I want to prove $$\begin{pmatrix} C & D \\ D & -C \end{pmatrix} \begin{pmatrix} A & B \\ -B & A \end{pmatrix} = \begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$ but I haven't had any luck yet, can you help?
 A: The $n\times n$ complex matrices $A+iB$ and $C+iD$ satisfy
$$
(A+iB)(C+iD)=(AC-BD)+i(BC+AD)=I
$$
so they are inverses of each other in $GL_n(\mathbb{C})$.  Hence
$$
I=(C+iD)(A+iB)=(CA-DB)+i(DA+CB)
$$
and you have the desired result by taking real and imaginary parts.
A: There were sign errors in the first version of your question. These are fixed now, but it is more convenient to write the hypothesis as
$$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & -D \\ D & C \end{pmatrix} =\begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, ,$$
and the desired conclusion as
$$\begin{pmatrix} C & -D \\ D & C \end{pmatrix} \begin{pmatrix} A & -B \\ B & A \end{pmatrix} = \begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, , $$
so that we have ($2n$-dimensional) unit matrices on the right-hand side.
These equations are equivalent because for square matrices, $XY = I$ is equivalent to $YX = I$, i.e. the left and right inverse of a matrix are the same.
Remark: The statement holds (and the proof works) not only for real-valued matrices, but more generally for complex-valued matrices, for matrices over any commutative ring, or even for elements of a finite-dimensional K-algebra. For more information, see

*

*If $AB = I$ then $BA = I$

*Right invertible and left zero divisor in matrix rings over a commutative ring
