If $A$ and $B$ are square matrices of the same order and $|B| \neq 0$,then $(B^{-1}AB)^5$ is equal to

Let $A = \begin{bmatrix} 2k - 1 & 1 & 1 \\ 0 & 2k - 1 & 1 \\ 0 & 0 & 2k - 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k - 1 & 1 \\ 1 - 2k & 0 & k \\ -1 & -k & 0 \end{bmatrix}$ where $k$ is a real number. If $\det(\text{adj } A) + \det(\text{adj } B) = 11^6$, then the value of $k - 5$ is equal to...

If $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$, then $A^{-1}$ is

If $A=\begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$,then $A-A^{-1}=$

If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$,then verify that $A \text{ adj } A = |A| I$. Also,find $A^{-1}$.

If $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$,then $A^{-1} =$