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 $2 \times 2$.

Which of the following matrices is invertible?
$A_{1}=\begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$
$A_{2}=\begin{bmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{bmatrix}$
$A_{3}=\begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{bmatrix}$
$A_{4}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$

If $A = \begin{bmatrix} k/2 & 0 & 0 \\ 0 & l/3 & 0 \\ 0 & 0 & m/4 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix}$,then $k+l+m=$

Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.

If $\begin{bmatrix} 5 & a & -7 \\ b & -7 & c \\ -7 & d & -1 \end{bmatrix}$ is the adjoint of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix}$,then $a+b+c+d=$

If $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \end{bmatrix}$,then the values of $a$ and $c$ are respectively: