If the system of equations  $x +y + z = 6$ ; $x + 2y + 3z= 10$ ; $x + 2y + \lambda z = 0$ has a unique solution, then $\lambda $ is not equal to

The value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{{\omega ^2}}&\omega \\1&\omega &{{\omega ^2}}\end{array}\,} \right| = $

The following system of linear equations  $7 x+6 y-2 z=0$ ; $3 x+4 y+2 z=0$ ; ${x}-2{y}-6{z}=0,$ has

If $A = \left| {\,\begin{array}{*{20}{c}}{ - 1}&2&4\\3&1&0\\{ - 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{*{20}{c}}{ - 2}&4&2\\6&2&0\\{ - 2}&4&8\end{array}\,} \right|$, then $B$ is given by

If $a, b, c$ are non-zero real numbers and if the system of equations $(a - 1 )x = y + z,$  $(b - 1 )y = z + x ,$ $(c - 1 )z= x + y,$ has a non-trivial solution, then $ab + bc + ca$ equals