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

The joint equation of pair of lines through the origin and making an equilateral triangle with the line $$y=3$$ is

If $$\bar{r}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$ is a linear combination of the vectors $$\bar{a}=-\hat{i}+4 \hat{j}+3 \hat{k}$$ and $$\bar{b}=-8 \hat{i}-\hat{j}+3 \hat{k}$$, then

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 $$(A B)^{-1}=$$