The system of equations $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ has no solution, if $\alpha $ is

The system of equations $\lambda x + y + z = 0,$ $ - x + \lambda y + z = 0,$ $ - x - y + \lambda z = 0$, will have a non zero solution if real values of $\lambda $ are given by

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 :

If ${A_\lambda } = \left( {\begin{array}{*{20}{c}}
\lambda &{\lambda  - 1}\\
{\lambda  - 1}&\lambda 
\end{array}} \right);\,\lambda  \in N$ then $|A_1| + |A_2| + ..... + |A_{300}|$ is equal to

The existence of the unique solution of the system $x + y + z = \lambda ,$ $5x - y + \mu z = 10$, $2x + 3y - z = 6$ depends on

Evaluate the determinants

$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$