If $$A^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5\end{array}\right]$$, then $$A=$$

If $$A^{-1}=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$$ and $$B^{-1}=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$$, then $$(A B)^{-1}=$$

$$A(\propto)=\left[\begin{array}{cc}\cos \propto & \sin \propto \\ -\sin \propto & \cos \propto\end{array}\right]$$, then $$\left[A^2(\propto)\right]^{-1}=$$

If inverse of $$\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$$ does not exist, then $$x=$$