Question

If $a+b+c>0$ and
$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ then

• A. $\Delta <0$
• B. $\Delta \le 0$
• C. $\Delta >0$
• D. $\Delta =0$

Answer 1

Answer: (B)

$\Delta \le 0$

Applying $C_1\to C_1+C_2+C_3$ we get

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

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ we get

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

Expanding along with $C_1$ we get

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

$=-(a+b+c)(b^2+c^2-2bc+a^2-ab-ac+bc)$

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

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

$=[\because a+b+c>0]$