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

Multiply ${C_1},\,{C_2},{C_3}$ by $a,\,\,b,\,c$ respectively and hence divide by $abc$

 $\Delta = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{a({a^2} + {x^2})}&{a{b^2}}&{{c^2}a}\\{{a^2}b}&{b({b^2} + {x^2})}&{b{c^2}}\\{c{a^2}}&{{b^2}c}&{c({c^2} + {x^2})}\end{array}\,} \right|$

Now take out $a, b $ and $c $ common from ${R_1},\,{R_2}$ and ${R_3}$,

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

Now applying ${C_1} \to {C_1} + {C_2} + {C_3}$

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

==> $\Delta = {x^4}({a^2} + {b^2} + {c^2} + {x^2})$

Hence, it is divisible by ${x^2}$.