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  • If a + b + c ≠ 0 and then prove that a = b = c.

If a + b + c ≠ 0 and \left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=0 then prove that a = b = c.

Answers (1)

\\ A=\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|

\\\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &=\left|\begin{array}{lll} a+b+c & b & c \\ b+a+c & c & a \\ c+a+b & a & b \end{array}\right|\\ &\text { Take }(a+b+c) \text { common from Column } 1\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right| \end{aligned}

Expand along Column 1

\\\vspace{\baselineskip} = (a + b + c)[(1)(bc -a\textsuperscript{2}) -(1)(b\textsuperscript{2} -ac) + (1)(ba -c\textsuperscript{2})]\\ \\ \vspace{\baselineskip}= (a + b + c)[bc -a\textsuperscript{2} -b\textsuperscript{2} + ac + ab -c\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= (a + b + c)[-(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2} -ab -bc -ac)]\\

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

Given that Δ = 0

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

\vspace{\baselineskip} \Rightarrow (a + b + c)[(a -b)^2 + (b -c)^2 + (c -a)^2] = 0\\ \vspace{\baselineskip} \text{Either } (a + b + c) = 0 \text{ or } (a -b)^2 + (b -c)^2 + (c -a)^2 = 0\\ \text{but it is given that } (a + b + c) \neq 0\\ \therefore (a -b)^2 + (b -c)^2 + (c -a)^2 = 0\\ \Rightarrow a -b = b -c = c -a = 0\\ \Rightarrow a = b = c

Hence Proved

 

 

Posted by

infoexpert22

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