If $$A=\left[\begin{array}{cc}2 & -1 \\ -1 & 3\end{array}\right]$$, then the inverse of $$\left(2 A^2+5 A\right)$$ is

If $$A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$$, then $$\operatorname{adj} A=$$

If $$A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1\end{array}\right]$$ and $$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc}1 & -1 & 1 \\ -8 & 6 & 2 c \\ 5 & -3 & 1\end{array}\right]$$, then values of a and c are respectively

If $$A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$$, then $$A^{-1}=$$