If $AB = A$ and $BA = B$, then

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

If $\omega $ is a cube root of unity and $\Delta = \left| {\begin{array}{*{20}{c}}1&{2\omega }\\\omega &{{\omega ^2}}\end{array}} \right|$, then ${\Delta ^2}$ is equal to

The value of a for which the system of equations ${a^3}x + {(a + 1)^3}y + {(a + 2)^3}z = 0,$ $ax + (a + 1)y + (a + 2)z = 0,$ $x + y + z = 0,$ has a non zero solution is

If $\omega $ be a complex cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}1&\omega &{ - {\omega ^2}/2}\\1&1&1\\1&{ - 1}&0\end{array}\,} \right| = $

If the system of equations  $2 x+3 y-z=5$  ;  $x+\alpha y+3 z=-4$  ;  $3 x-y+\beta z=7$ has infinitely many solutions, then $13 \alpha \beta$ is equal to