If $f: N \to Z$ is defined by $f(n) = \det \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3(2k+1) & 3k+7 \end{vmatrix}$, where $k \in N$ and $\sum_{n=1}^k f(n) = 98$, then $k$ 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:

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

Let $M$ and $m$ respectively be the maximum and the minimum values of $f(x) = \left| \begin{array}{ccc} 1+\sin^2 x & \cos^2 x & 4\sin 4x \\ \sin^2 x & 1+\cos^2 x & 4\sin 4x \\ \sin^2 x & \cos^2 x & 1+4\sin 4x \end{array} \right|$,$x \in R$. Then $M^4 - m^4$ is equal to:

The rank of $\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1\end{array}\right]$ is

If $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092} \end{bmatrix}$,then the rank of $A$ is