Question

If $\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ is a singular matrix and $a,b,c$ are positive, then

• A. $a+b+c=0$
• B. $a=b=c$
• C. $a^2+b^2+c^2+3abc=0$
• D. $a^2+b^2+c^2-3abc\ne 0$

Answer 1

The correct option is B $a=b=c$

For a singular matrix

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

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

$\Rightarrow abc-a^3-b^3+abc+abc-c^3=0$

$\Rightarrow a^3+b^3+c^3-3abc=0$

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

$\Rightarrow (a+b+c)\ne 0$ $because$ a,b,c are $+ve$ $numbers$

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

$\Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=0\times 2$

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

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

it is possible if and only if $a=b=c$