Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :

The solutions of the equation $\left| {\,\begin{array}{*{20}{c}}x&2&{ - 1}\\2&5&x\\{ - 1}&2&x\end{array}\,} \right| = 0$ are

Evaluate the determinants : $\left|\begin{array}{ll}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|$

$\Delta = \left| {\,\begin{array}{*{20}{c}}{a + x}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right|$,which of the following is a factor for the above determinant

If $\left| {\,\begin{array}{*{20}{c}}{1 + ax}&{1 + bx}&{1 + cx}\\{1 + {a_1}x}&{1 + {b_1}x}&{1 + {c_1}x}\\{1 + {a_2}x}&{1 + {b_2}x}&{1 + {c_2}x}\end{array}\,} \right|,$ $ = {A_0} + {A_1}x + {A_2}{x^2} + {A_3}{x^3}$ then ${A_1}$ is equal to

Let the area of the triangle with vertices $A (1, \alpha)$, $B (\alpha, 0)$ and $C (0, \alpha)$ be $4\, sq.$ units. If the point $(\alpha,-\alpha),(-\alpha, \alpha)$ and $\left(\alpha^{2}, \beta\right)$ are collinear, then $\beta$ is equal to