The positive value of the determinant of the matrix $A$,whose $\operatorname{Adj}(\operatorname{Adj}(A)) = \begin{bmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{bmatrix}$,is

The set of all $2 \times 2$ matrices over the real numbers is not a group under matrix multiplication because

If $A$ is an invertible matrix of order $n$, then the determinant of $\operatorname{adj} A$ is equal to :

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$,$10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$ and $B$ is the inverse of $A$,then the value of $\alpha$ is

If $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right],$ then verify that $(AB)^{-1}=B^{-1} A^{-1}$.

If $A = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1} A^{-1})^{-1} = $