If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then $A^{-1}$ is given by

Find the inverse of each of the matrices (if it exists). $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A-3 I)(A-5 I)=O$,where $I=I_3$ and $O=O_3$. Here $O_3$ represents the zero matrix of order $3$ and $I_3$ is the identity matrix of order $3$. If $\alpha A+\beta A^{-1}=4 I$,then $\alpha+\beta$ is equal to:

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 $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}$,then the values of $\alpha$ and $\beta$ are respectively:

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \text{adj}(A) = AA^T$,then $5a + b =$