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

If $A=\begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A A^{T}$,then $2a + 3b$ is

If $A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$,then $x =$ . . . . . . .

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ such that $A^2 - 4A + 3I = 0$,where $I$ is a unit matrix of order $2$,then $A^{-1}$ is

If $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$ and $PQ = I$,then find $(\csc^2 \theta)Q$,where $I$ is an identity matrix of order $2 \times 2$.

For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$,show that $A^{3} - 6A^{2} + 5A + 11I = 0$. Hence,find $A^{-1}$.