સાબિત કરો કે, $\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-x \mathrm{R}_{2}$ to $\Delta,$ we get
${\Delta = \left| {\begin{array}{*{20}{c}}
{a\left( {1 - {x^2}} \right)}&{c\left( {1 - {x^2}} \right)}&{p\left( {1 - {x^2}} \right)} \\
{ax + b}&{cx + d}&{px + q} \\
u&v&w
\end{array}} \right|}$
${ = \left( {1 - {x^2}} \right)\left| {\begin{array}{*{20}{c}}
a&c&p \\
{ax + b}&{cx + d}&{px + q} \\
u&v&w
\end{array}} \right|}$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-x \mathrm{R}_{1},$ we get
$\Delta=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right|$
જો $x, y, z$ ભિન્ન હોય અને $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0$ હોય, તો સાબિત કરો કે $1+x y z=0$.
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી સાબિત કરો કે, $\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c - a}&{c + a - b}&{a + b - c}\end{array}\,} \right|$ = . .
$\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ ની કિમત મેળવો.