Question

If $a+b+c\ne 0$ and $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$, then using properties of determinants, prove that $a=b=c$.

Answer 1

$△=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

$\Rightarrow C_1=C_1+C_2+C_3$

$\Rightarrow △=\left|\begin{array}{ccc}a+b+c& b& c\\ a+b+c& c& a\\ a+b+c& a& b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|=0$

$\Rightarrow R_2=R_2-R_1;R_3=R_3-R_1$

$\Rightarrow (a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& c-b& a-c\\ 0& a-b& b-c\end{array}\right|=0$

expanding along $C_1$

$\Rightarrow (a+b+c)\times 1\left|\begin{array}{cc}c-b& a-c\\ a-b& b-c\end{array}\right|=(a+b+c)[(c-b)(b-c)-(a-b)(a-c)]$

$\Rightarrow (a+b+c)[bc-b^2+bc-c^2-a^2+ab-c^2-bc+ac]=0$

$\Rightarrow -(a+b+c)[a^2+b^2+c^2-ab-bc-ac]=0$

$\frac{-1}{2}(a+b+c)[{2a}^2+2b^2+{2c}^2-2ab-2bc-2ac]$

$\Rightarrow \frac{-1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$

$\because (a+b+c)\ne 0$

$\therefore {(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$

This can be true only when

$(a-b)=0,$ $(b-c)=0,$ $(c-a)=0$

$\Rightarrow a=b,b=c,c=a$

$\therefore a=b=c.$

Hence, proved.