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If $\mathbf{a}, \mathbf{b}$ 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 $a=b=c$.
$\Delta=\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=0,$
Show that either $a+b+c=0$ or $a=b=c$.
Solution:
2461 Upvotes
Verified Answer
$\Delta=\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|$
Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$
$\Delta=\left|\begin{array}{lll}2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ 2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{a}+\mathrm{b} & \mathrm{b}+\mathrm{c} \\ 2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a}\end{array}\right|$
$\Rightarrow \Delta=2\left|\begin{array}{lll}\mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b} & \mathrm{b}+\mathrm{c} \\ \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a}\end{array}\right|$
Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1$
$\Delta=2\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{b} & -\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{c} & -\mathrm{a} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{a} & -\mathrm{b}
\end{array}\right|$
Taking $(a+b+c)$ common from $C_1,(-1)$ from $C_2$ and $C_3$.
$\Delta=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & \mathrm{~b} & \mathrm{c} \\
1 & \mathrm{c} & \mathrm{a} \\
1 & \mathrm{a} & \mathrm{b}
\end{array}\right|$
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|$
Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$
$\Delta=\left|\begin{array}{lll}2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ 2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{a}+\mathrm{b} & \mathrm{b}+\mathrm{c} \\ 2(\mathrm{a}+\mathrm{b}+\mathrm{c}) & \mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a}\end{array}\right|$
$\Rightarrow \Delta=2\left|\begin{array}{lll}\mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b} & \mathrm{b}+\mathrm{c} \\ \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a}\end{array}\right|$
Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1$
$\Delta=2\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{b} & -\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{c} & -\mathrm{a} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{a} & -\mathrm{b}
\end{array}\right|$
Taking $(a+b+c)$ common from $C_1,(-1)$ from $C_2$ and $C_3$.
$\Delta=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & \mathrm{~b} & \mathrm{c} \\
1 & \mathrm{c} & \mathrm{a} \\
1 & \mathrm{a} & \mathrm{b}
\end{array}\right|$
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