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

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Question

(A) Let $\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)}

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) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight|$.Applying the operation $R_2 	o R_2 - R_3$:Since $(x+1)^2 - (x-1)^2 = 4x$,the determinant becomes:$\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\4a&4b&4c\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight| = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\a&b&c\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} ight|$.Now,apply $R_3 	o R_3 - R_1 + 2R_2$:Note that $(x-1)^2 - x^2 + 2x = x^2 - 2x + 1 - x^2 + 2x = 1$.Thus,the third row becomes $1, 1, 1$.Therefore,$\Delta = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\a&b&c\1&1&1\end{array}} ight|$.Hence,the correct option is $A$.

$\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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