If a, b and c are real numbers, and

Question:

If $a, b$ and $c$ are real numbers, and $\Delta=\left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$,

Show that either a + b + c = 0 or a = b = c.

Solution:

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

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have:

$\Delta=\left|\begin{array}{lcc}2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|$

$=2(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

$\Delta=2(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b\end{array}\right|$

Expanding along $R_{1}$, we have:

$\begin{aligned} \Delta &=2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)] \\ &=2(a+b+c)\left[-b^{2}-c^{2}+2 b c-b c+b a+a c-a^{2}\right] \\ &=2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right] \end{aligned}$

It is given that $\Delta=0$.

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

$\Rightarrow$ Either $a+b+c=0$, or $a b+b c+c a-a^{2}-b^{2}-c^{2}=0$.

Now,

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

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

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

$\Rightarrow(a-b)^{2}=(b-c)^{2}=(c-a)^{2}=0 \quad\left[(a-b)^{2},(b-c)^{2},(c-a)^{2}\right.$ are non-negative $]$

$\Rightarrow(a-b)=(b-c)=(c-a)=0$

$\Rightarrow a=b=c$

Hence, if $\Delta=0$, then either $a+b+c=0$ or $a=b=c$.

 

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