Let a, b, c 0 and Delta = begin{vmatrix} a+ | vedclass

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(C) Given $\Delta = \begin{vmatrix} a+b & b & c \ b+c & c & a \ c+a & a & b \end{vmatrix}$.Applying $C_1 ightarrow C_1 - C_2$,we get $\Delta = \be

Vedclass · Accepted Answer

$\Delta = 0 \Rightarrow a + b + c = 0$

Vedclass · Answer

(C) Given $\Delta = \begin{vmatrix} a+b & b & c \ b+c & c & a \ c+a & a & b \end{vmatrix}$.Applying $C_1 ightarrow C_1 - C_2$,we get $\Delta = \begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}$.Expanding the determinant,we get $\Delta = a(bc - a^2) - b(b^2 - ac) + c(ab - c^2) = abc - a^3 - b^3 + abc + abc - c^3 = -(a^3 + b^3 + c^3 - 3abc)$.This can also be written as $\Delta = -\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]$.Since $a, b, c > 0$,we have $a+b+c > 0$. Also,the sum of squares is always non-negative. Thus,$\Delta \leq 0$.If $\Delta = 0$,then either $a+b+c = 0$ (which is impossible as $a, b, c > 0$) or $(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$,which implies $a=b=c$.Therefore,the statement $\Delta = 0 \Rightarrow a+b+c = 0$ is incorrect.

Let $a, b, c > 0$ and $\Delta = \begin{vmatrix} a+b & b & c \\ b+c & c & a \\ c+a & a & b \end{vmatrix}$. Then which of the following is not correct?

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