The area of a triangle is $4$ sq. units,and its vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$. Find the value of $k$.

If $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,then for all $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$,$\det(A)$ lies in the interval:

If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

Evaluate the determinant: $\left| \begin{array}{ccc} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{array} \right|$

The system of linear equations $x + \lambda y - z = 0, \lambda x - y - z = 0, x + y - \lambda z = 0$ has a non-trivial solution for: