Relatiosn with Matrices

Question

For every integer $n>2$ and $A,B,C,D \in M_{2}(\mathbb{R})$ we have :

$1) AC-BD=I_{n}$

$2) AD+BC=O_{n}$

Prove that $CA-DB=I_{n}$ and $DA+CB=O_{n}$.

I tried to prove that $AC=CA$ like the proof of theorem " $AB=I$ , then $BA=I$ " , after that I tried with multiply and add the relations but I didn't manage to prove that. Any idea?

Answer 1

Consider the block multiplication $$ \begin{bmatrix} A & -B \\ B & A \end{bmatrix} \begin{bmatrix} C & - D \\ D & C \end{bmatrix}= \begin{bmatrix} AC-BD & -(AD+BC) \\ BC+AD & -BD+AC \end{bmatrix} =\begin{bmatrix} I_n & O_n \\ O_n & I_n \end{bmatrix}=I_{2n} $$ Then also $$ \begin{bmatrix} C & - D \\ D & C \end{bmatrix} \begin{bmatrix} A & -B \\ B & A \end{bmatrix} =I_{2n} $$

Answer 2

To prove that determinant of AC is positive, we solve like that or we use the relations. I try like that:

$\det(AC)=\det(BD+I_{n})=?$

I thought that $\det(BD+I_{n})=1+\mathrm{Tr}(BD)+\det(BD)$.