For a non - zero, real a, b and c | * 20 c a^2 + b^2 c &c&c & b^2… - Vidyadip

MCQ

For a non - zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

A
$-4$
B
$0$
C
$2$
✓
$4$

✓

Answer

Correct option: D.

$4$

d
$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{{c^2}}&{{c^2}}\\ {{a^2}}&{{b^2} + {c^2}}&{{a^2}}\\{{b^2}}&{{b^2}}&{{c^2} + {a^2}}\end{array}\,\,} \right|$

use $R_1 \rightarrow R_1 - (R_2 + R_3)$

$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}0&{ - 2{b^2}}&{ - 2{a^2}}\\{{a^2}}&{{b^2} + {c^2}}&{{a^2}}\\{{b^2}}&{{b^2}}&{{c^2} + {a^2}} \end{array}\,\,} \right|$

$R_2 \rightarrow R_2 + 1/2R_1$ and $R_3 \rightarrow R_3 + 1/2 R_1$

$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}0&{ - 2{b^2}}&{ - 2{a^2}}\\ {{a^2}}&{{c^2}}&0\\{{b^2}}&0&{{c^2}}\end{array}\,\,} \right|$
$\frac{1}{{abc}}\,$ [$ 2b^2 (a^2 c^2) - 2a^2 (- b^2 c^2) $] =$\frac{{4{a^2}\,{b^2}\,{c^2}}}{{abc}}\, = \,4abc$

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