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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
If $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}\,} \right| = k\,abc{(a + b + c)^3}$, then the value of $k$ is
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Answer

c
(c) Operate ${C_2} \to {C_2} - {C_1},\,{C_3} \to {C_3} - {C_1}$ and take out $a + b + c$ from ${C_2}$ as well as from ${C_3}$ to get

$\Delta = {(a + b + c)^2}$ $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{a - b - c}&{a - b - c}\\{{b^2}}&{c + a - b}&0\\{{c^2}}&0&{a + b - c}\end{array}\,} \right|$

(Operate ${R_1} \to {R_1} - {R_2} - {R_3}$)

= ${(a + b + c)^2}\,\left| {\,\begin{array}{*{20}{c}}{2bc}&{ - 2c}&{ - 2b}\\{{b^2}}&{c + a - b}&0\\{{c^2}}&0&{a + b - c}\end{array}\,} \right|$

(Operate ${C_2} \to {C_2} + \frac{1}{b}{C_1}$ and ${C_3} \to {C_3} + \frac{1}{c}{C_1})$

= ${(a + b + c)^2}\left| {\begin{array}{*{20}{c}}{2bc}&0&0\\{{b^2}}&{c + a}&{\frac{{{b^2}}}{c}}\\{{c^2}}&{\frac{{{c^2}}}{b}}&{a + b}\end{array}} \right|$

$ = \,{(a + b + c)^2}[2bc\{ (a + b)\,(c + a) - bc\} ]$

$ = \,2abc{(a + b + c)^3}$.

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