Question

Let $\Delta =\left|\begin{array}{ccc}\sum a^2& \sum ab& \sum ab\\ \sum ab& \sum a^2& \sum ab\\ \sum ab& \sum ab& \sum a^2\end{array}\right|$
Prove that $\Delta$ is non-negative and establish the relation between $a,b$ and $c$ if $\Delta =0.$

Answer 1

Apply $C_1+C_2+C_3$ and $\sum a^2+2\sum ab={(\sum a)}^2$ be

taken out from new $C_1$.

$\Delta ={(\sum a)}^2\left|\begin{array}{ccc}1& \sum ab& \sum ab\\ 1& \sum a^2& \sum ab\\ 1& \sum ab& \sum a^2\end{array}\right|$

Now making two zeros by $R_2-R_1$ and $R_3-R_1$

$\Delta ={(\sum a)}^2{[\sum a^2-\sum ab]}^2$

$\Delta$ being perfect square of real quantities is always $+$ive.

$\Delta =0$ if either $\sum a=0$ or $\sum a^2-\sum ab=0$

i.e. $1╱2[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$

Above will hold only when $a-b=0$, $b-c=0$, $c-a=0$ or $a=b=c$.