If a matrix $A$ satisfies the equation $A^3-6A^2+11A-6I=0$,then $A^{-1}$ can be expressed in terms of $A$ as:

The inverse of the matrix $ \begin{bmatrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{bmatrix} $ is

If $A$ is a $3 \times 3$ matrix and $|A|=2$,then $|\operatorname{Adj}(\operatorname{Adj} A)| \operatorname{Adj}(\operatorname{Adj} A) = $ (in $A$)

Let $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\3&2&1\end{array}} \right]$. If $u_1$ and $u_2$ are column matrices such that $A{u_1} = \left[ {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right]$ and $A{u_2} = \left[ {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right]$,then $u_1 + u_2$ is equal to:

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

Let $A$ be an invertible square matrix of order $3 \times 3$. Then $|(\text{adj} A) \cdot A|$ is