જો $\left| {\,\begin{array}{*{20}{c}}{x - 1}&3&0\\2&{x - 3}&4\\3&5&6\end{array}\,} \right| = 0$ તો $x =$

$\left| {{\rm{ }}\begin{array}{*{20}{c}}1&2&3\\3&5&7\\8&{14}&{20}\end{array}} \right| = . . . $

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

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}{x - 1}&1&1\\1&{x - 1}&1\\1&1&{x - 1}\end{array}\,} \right| = 0$  ના બીજ મેળવો.

સમીકરણની સંહતિ ${x_1} - {x_2} + {x_3} = 2,$ $\,3{x_1} - {x_2} + 2{x_3} = - 6$ અને $3{x_1} + {x_2} + {x_3} = - 18$ નો ઉકેલ . . . .

જો $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}\,} \right| = 0$; તો $a,b,c$ એ .. . . શ્રેણીમાં છે .