The equation $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{ - \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 - 5x}\\{x + 1}&{2x}&{2 - 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 - x)}^2}}&{3x}&{2x}\\{1 - 2x}&{3x - 2}&{2x - 3}\end{array}} \right|$ $= 0$

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

If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ then $\lambda $ is

$x + ky - z = 0,3x - ky - z = 0$ and $x - 3y + z = 0$ has non-zero solution for $k =$

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to

If the system of linear equations $x + 2ay + az = 0$ $x + 3by + bz = 0$ $x + 4cy + cz = 0$ has a non-zero solution, then $a, b, c$