left| {begin{array}{*{20}{c}}{{a^2}}&{{b^2}}& | vedclass

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(A) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\{{{(a - 1)}^2}}&{{{(b -

Vedclass · Accepted Answer

$4\,\left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\a&b&c\1&1&1\end{array}} ight|$

Vedclass · Answer

(A) ધારો કે $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight|$.પ્રક્રિયા $R_2 	o R_2 - R_3$ લાગુ કરતા:કારણ કે $(x+1)^2 - (x-1)^2 = 4x$,નિશ્ચાયક આ મુજબ બનશે:$\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\4a&4b&4c\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight| = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\a&b&c\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight|$.હવે,$R_3 	o R_3 - R_1 + 2R_2$ લાગુ કરતા:નોંધો કે $(x-1)^2 - x^2 + 2x = x^2 - 2x + 1 - x^2 + 2x = 1$.આમ,ત્રીજી હાર $1, 1, 1$ બને છે.તેથી,$\Delta = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\a&b&c\1&1&1\end{array}} ight|$.આમ,સાચો વિકલ્પ $A$ છે.

$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right| = $

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