67th Romanian Mathematical Olympiad

$$(CA+kDB)-w(DA-CB)=I_n$$ and $$(CA+kDB)+w(DA-CB)=I_n,$$ which give the result. For $k=0$ we get $AC=I_n$, so $CA=I_n$. From $AD=BC$ and $CA=I_n$ we obtain $ADA=B$ and $DA=CB$.