The rank of the matrix $\begin{bmatrix} 4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x \end{bmatrix}$ is $1$,then

If $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$, then the value of $\sum_{r=1}^{n} S_{r}$ is independent of

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:

The rank of the matrix $A=\begin{bmatrix} 1 & -1 & 0 & -2 \\ -4 & 4 & 0 & 8 \\ -2 & 1 & 2 & 4 \end{bmatrix}$ is

The value of the determinant $\left| \begin{array}{ccc} a^2 & a & 1 \\ \cos(nx) & \cos(n+1)x & \cos(n+2)x \\ \sin(nx) & \sin(n+1)x & \sin(n+2)x \end{array} \right|$ is independent of :

Let $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$. If $r$ is the rank of $A$, then $r + x =$