If $1,\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\\{{\omega ^{2n}}}&1&{{\omega ^n}}\end{array}\,} \right|$ is equal to
$0$
$1$
$\omega $
${\omega ^2}$
જો ${\Delta _r} = \left| {\begin{array}{*{20}{c}}
r&{2r - 1}&{3r - 2} \\
{\frac{n}{2}}&{n - 1}&a \\
{\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)}
\end{array}} \right|$ તો $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $ ની કિમત . . .
જો ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}x&3\\{ - 2}&1\end{array}\,} \right|$ તો $x =$
$(3, 8), (-4, 2)$ અને $(5, 1)$ શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ શોધો.
જો $\left| {{\kern 1pt} \begin{array}{*{20}{c}}1&2&3\\2&x&3\\3&4&5\end{array}\,} \right| = 0 $ તો $ x =$