The rank of the matrix $\left[\begin{array}{cccc}3 & 2 & 1 & -4 \\ 2 & 3 & 0 & -1 \\ 1 & -6 & 3 & -8\end{array}\right]$ is

The determinant $\left| \begin{array}{ccc} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos 2\phi \\ \sin \theta & \cos \theta & \sin \phi \\ -\cos \theta & \sin \theta & \cos \phi \end{array} \right|$ is :

If $A = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$ and $B = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix}$,then $\frac{dA}{dx}$ is equal to

If $A = \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 8 & 12 \\ 0 & 0 & 0 & 4 & 8 \end{bmatrix}$,then the rank of $A$ is

If $a_1, a_2, a_3, \dots, a_n$ form a geometric progression,find the value of the determinant: $\left| \begin{array}{ccc} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array} \right|$.

If $f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x \end{array} \right|$,then $\lim_{x \rightarrow \pi} f(x)$ is equal to