If a, b and c are real numbers, and = b+c&c+a&a+b +a&a+b&b+c +b&b… - Vidyadip

Question

If $a, b$ and $c$ are real numbers, and $\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,$ Show that either $a + b + c = 0$ or $a = b = c.$

✓

Answer

$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
Applying $R_{1 }\rightarrow R_1 + R_2 + R_3,$ we have:
$\triangle=\begin{vmatrix}2(a+b+c)&2(a+b+c)&2(a+b+c)\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&1&1\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&0&0\\c+a&a-b&b-a\\a+b&c-a&c-b\end{vmatrix}\ \text{C}_1\rightarrow\text{C}_2-\text{C}_1\ \text{and}\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
Expanding along $R_1$, we have:
$\triangle = 2(a + b + c) (1) [(b - c) (c - b) - (b - a) (c - a)]$
$= 2(a + b + c) [-b^{2 }- c^2 + 2bc - bc + ba + ac - a^2]$
$= 2(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2]$
It is given that $\triangle = 0.$
$(a + b + c) [ab + bc + ca - a^2 - b^2 - c^2] = 0$
$\Rightarrow$ Either $a + b + c = 0,$ or $ab + bc + ca - a^2 - b^2 - c^2 = 0.$
Now,
$ab + bc + ca - a^2 - b^2 - c^2 = 0$
$\Rightarrow -2ab - 2bc - 2ca + 2a^2 + 2b^2 + 2c^2 = 0$
$\Rightarrow (a - b)^2 + (b - c)^2 + (c - a)^2 = 0$
$\Rightarrow (a -b)^{2 }= (b - c)^2 = (c - a)^2 = 0 [(a - b)^2, (b - c)^2, (c - a)^2$ are non$-$negative$]$
$\Rightarrow (a - b) = (b - c) = (c - a) = 0$
$\Rightarrow a = b = c$
Hence, if $\triangle = 0,$ then either $a + b + c = 0$ or $a = b = c.$

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