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

If the system of equations $a_1 x + b_1 y + c_1 z = 0$,$a_2 x + b_2 y + c_2 z = 0$,and $a_3 x + b_3 y + c_3 z = 0$ has only the trivial solution,then the rank of the matrix $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$ is:

If the points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ are collinear,then the rank of the matrix $\begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix}$ will always be less than

If $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ and the rank of $A$ is $2$,then the value of $x$ is equal to

For $\triangle ABC$,evaluate the determinant: $\left|\begin{array}{ccc}0 & \sin A & \tan B \\ -\sin ( B + C ) & 0 & \cos C \\ \tan ( A + C ) & -\cos C & 0\end{array}\right|=$ . . . . . . .

Let $A=\left[\begin{array}{rrr}-1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]$,$B=\left[\begin{array}{rr}1 & -2 \\ -1 & 2\end{array}\right]$ and $C=\left[\begin{array}{rrr}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$. If $a, b$ and $c$ respectively denote the ranks of $A, B$ and $C$,then the correct order of these numbers is: