The rank of the matrix $\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 1 & -2 \\ 1 & -1 & 4 \\ 2 & 2 & 8\end{array}\right]$ is

If $f'(x) = \left| \begin{array}{ccc} mx & mx - p & mx + p \\ n & n + p & n - p \\ mx + 2n & mx + 2n + p & mx + 2n - p \end{array} \right|$,then $y = f(x)$ represents

Let $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$. If $r$ is the rank of $A$, then $r + x =$

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:

Let $f(x) = \left|\begin{array}{ccc} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{array}\right|$,where $a \in R$. Then the sum of the squares of all the values of $a$ for which $2f'(10) - f'(5) + 100 = 0$ is:

For some $a, b$,let $f(x) = \left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{array}\right|, \quad x \neq 0$. If $\lim_{x \rightarrow 0} f(x) = \lambda + \mu a + \nu b$,then $(\lambda + \mu + \nu)^2$ is equal to: