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 $\Delta(x) = \begin{vmatrix} x-2 & (x-1)^2 & x^3 \\ x-1 & x^2 & (x+1)^3 \\ x & (x+1)^2 & (x+2)^3 \end{vmatrix}$, then the coefficient of $x$ in $\Delta(x)$ is:

If $\left| \begin{array}{ccc} 1 + ax & 1 + bx & 1 + cx \\ 1 + a_1x & 1 + b_1x & 1 + c_1x \\ 1 + a_2x & 1 + b_2x & 1 + c_2x \end{array} \right| = A_0 + A_1x + A_2x^2 + A_3x^3$,then $A_1$ is equal to:

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

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 $r + x =$

If $\alpha, \beta, \text{ and } \gamma$ are real numbers,then $D = \begin{vmatrix} 1 & \cos(\beta - \alpha) & \cos(\gamma - \alpha) \\ \cos(\alpha - \beta) & 1 & \cos(\gamma - \beta) \\ \cos(\alpha - \gamma) & \cos(\beta - \gamma) & 1 \end{vmatrix} = $