The rank of the matrix $\begin{bmatrix} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{bmatrix}$ is

Suppose $\alpha, \beta, \gamma$ are the roots of the equation $x^3+qx+r=0$ (where $r \neq 0$) and they are in $A$.$P$. Then the rank of the matrix $\begin{bmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{bmatrix}$ is

$A$ 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:

Let $f(x) = \left| \begin{array}{ccc} 1 + \sin^2 x & \cos^2 x & 4 \sin 2x \\ \sin^2 x & 1 + \cos^2 x & 4 \sin 2x \\ \sin^2 x & \cos^2 x & 1 + 4 \sin 2x \end{array} \right|$,then the maximum value of $f(x)$ is:

The determinant $\left| {\begin{array}{ccc} 4 + {x^2} & -6 & -2 \\ -6 & 9 + {x^2} & 3 \\ -2 & 3 & 1 + {x^2} \end{array}} \right|$ is not divisible by

If $f(x) = \left| \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & x \end{array} \right|$,then $\lim_{x \to 0} \frac{f(x)}{x^2}$ is equal to