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

If the vectors $\vec{\alpha}=\hat{i}+a \hat{j}+a^{2} \hat{k}$, $\vec{\beta}=\hat{i}+b \hat{j}+b^{2} \hat{k}$, and $\vec{\gamma}=\hat{i}+c \hat{j}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$, then the value of $abc$ is

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

The determinant $\left| \begin{array}{ccc} ^x{C_1} & ^x{C_2} & ^x{C_3} \\ ^y{C_1} & ^y{C_2} & ^y{C_3} \\ ^z{C_1} & ^z{C_2} & ^z{C_3} \end{array} \right|$ equals:

Let $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{array} \right|$. Then, find the value of $\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$.

Let $f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix}$. Then $\lim_{x \to 0} \frac{f'(x)}{x} =$