The value of the determinant$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 - x}&1\\1&1&{1 + y}\end{array}\,} \right|$is
$3 - x + y$
$(1 - x)(1 + y)$
$xy$
$ - xy$
If ${\Delta _r} = \left| {\begin{array}{*{20}{c}}
r&{2r - 1}&{3r - 2} \\
{\frac{n}{2}}&{n - 1}&a \\
{\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)}
\end{array}} \right|$ then the value of $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $
The system of equations $-k x+3 y-14 z=25$ $-15 x+4 y-k z=3$ $-4 x+y+3 z=4$ is consistent for all $k$ in the set
Consider the system of linear equations ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0$, ${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$ and ${a_3}x + {b_3}y + {c_3}z + {d_3} = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ if $\Delta (a,b,c) \ne 0$, then the value of $x$ in the unique solution of the above equations is
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