Find the rank of the matrix $\begin{bmatrix} 1 & 4 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{bmatrix}$.

The rank of the matrix $\begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & a - 4 \\ 1 & -2 & a + 1 \end{bmatrix}$ is

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

If $\alpha, \beta, \text{ and } \gamma$ are real numbers,then $D = \begin{vmatrix} 1 & \cos(\beta - \alpha) & \cos(\gamma - \alpha) \\ \cos(\alpha - \beta) & 1 & \cos(\gamma - \beta) \\ \cos(\alpha - \gamma) & \cos(\beta - \gamma) & 1 \end{vmatrix} = $

If $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right|$,then find $\lim_{x \to 0} \frac{f'(x)}{x}$.

$A$ is an $m \times n$ matrix of rank $4$. If $A$ contains an $m$-th order non-singular submatrix and $A^T A$ is a $7 \times 7$ matrix, then the number of rows of $A$ is: