જો $\left| {\begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\ 
  {{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\ 
  {{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}} 
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\
  a&b&c \\
  1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ તો $k$ મેળવો.

ધારો કે  $a-2 b+c=1$ છે . જો $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ હોય તો  . . . 

જો $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ તો $x$ મેળવો.

$\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right| = . . $

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

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