If $f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \\ x - a & x^2 - x & c + x \\ x - b & x - c & 0 \end{array} \right|$,then:

If $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$,then $\frac{1}{5} f'(0)$ is equal to:

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

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

The trace of a square matrix is defined as the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18$,then the value of the determinant of $A$ is:

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: