$\left| {\,\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}\,} \right| = $

$\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ નું મૂલ્ય ............ છે.

જો $\alpha $, $\beta$ $\gamma$, $\delta$ એ  $z^5=1$ ના કાલ્પનિક બીજ હોય તો  $\left| {\begin{array}{*{20}{c}}
  {{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha  + 1}}}&{ - {e^{ - \delta }}} \\ 
  {{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta  + 1}}}&{ - {e^{ - \delta }}} \\ 
  {{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma  + 1}}}&{ - {e^{ - \delta }}} 
\end{array}} \right|$ મેળવો.

સાબિત કરો કે, $\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$

$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $

 $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\{\cos (p - d)x}&{\cos px}&{\cos (p + d)x}\\{\sin (p - d)x}&{\sin px}&{\sin (p + d)x}\end{array}\,} \right|$ ની કિમંત . . .  પર આધારિત નથી.