For every $ n> 2, n\epsilon \mathbb{N}$ let A,B,C,D be matices $ A,B,C,D\epsilon M_{n}(\mathbb{R})$ such that $ AC-BD=I_{n}$ and $AD+BC=0_{n} $. Prove that:
a) $CA-DB=I_{n}$ and $DA+CB=0_{n}$
b)$ det(AC)\geq 0 $.
My postgraduate degree professor gave us this question. For a) i did the following $$ AC-BD=I_{n} \Leftrightarrow \\ \sum_{s=i}^{n}a_{is}c_{sj}-b_{is}d_{sj}=0 \forall i\neq j \text{ and } \sum_{s=i}^{n}a_{is}c_{sj}-b_{is}d_{sj}=1 \text{ if } i=j \Leftrightarrow \\\sum_{s=i}^{n}c_{sj}a_{is}-d_{sj}b_{sj}=0 \forall i\neq j \text{ and } \sum_{s=i}^{n}c_{sj}a_{is}-d_{sj}b_{is}=1 \text{ if } i=j \Leftrightarrow \\ CA-DB=I_{n} \text{ and } AD+BC=0_{n} \Leftrightarrow \\ \sum_{s=1}^{n}a_{is}d_{sj}+b_{is}c_{sj}=0 \text{ for } s=1,...,n \Leftrightarrow \\ \sum_{s=1}^{n}d_{sj}a_{sj}+c_{sj}b_{sj}=0 \text{ for } s=1,...,n \Leftrightarrow \\ DA+CB=0_{n} $$ and but it seems like something is missing. For b i thought of using $det(AC)=det(A)\cdot det(B)$ but either than this i am lost. Please help me.
for b) because $AD+BC=0_{n}$ we get $det(AC+iBC)=det(AC-iAD)$ so $detC=det[(C+iD)(A+iB)C]=det(C+iD)det(AC+iBC)=det(C+iD)det(AC-iAD)=det[(C+iD)A(C-iD)]=detA|det(C+iD)|^{2}$ so $detAC=(detA)^{2}|det(C+iD)|^{2}\geq 0$