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, $a=b=c$.

• A. True
• B. False

Answer 1

The correct option is A True

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

Applying $R_1\to R_1+R_2+R_3$, we get,

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

Taking common $(a+b+c)$ from $R_1$, we get,

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

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, we get,

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

On expanding along $R_1$, we get,

$A=(a+b+c)\left[\right(c-b\left)\right(b-c)-(a-b\left)\right(a-c\left)\right]$

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

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

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

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

$A=-\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

Since, $A=0$, therefore,

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

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

$a-b=b-c=c-a=0$

$a=b=c$