Let $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & k-1 \\ 0 & 0 & k-1 & 1 \end{bmatrix}$ and $k \in R$. Then,the value of $k$,if it exists,for which the rank of $A$ is $2$,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

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

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

If $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$,then the ratio $f^{\prime \prime}(x) : f^{\prime}(x) =$

The rank of the matrix $\left[ {\begin{array}{*{20}{c}}4&1&0&0\\3&0&1&0\\6&0&2&0\end{array}} \right]$ is: