If $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$,then $|\operatorname{adj}(\operatorname{adj}(2A))|$ is equal to:

If $A = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}$,then $A + \operatorname{adj}(A)$ is:

Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$. If $B$ is the inverse of matrix $A$,then $\alpha$ is

Inverse of the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \\ -c+id & a-ib\end{array}\right]$, then $A^{-1}$ is equal to

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$