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If $a \neq b \neq c$ are all positive, then the value of the determinant
$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is
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$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is
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Verified Answer
The correct answer is:
negative
$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=\left|\begin{array}{lll}a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b\end{array}\right|$
$\left(\right.$ Applying $\left.\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right)$
$=(a+b+c)\left|\begin{array}{lll}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{array}\right|$
[on taking $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\left.\mathrm{C}_{1}\right]$
$=(a+b+c)\left[1\left(b c-a^{2}\right)-b(b-a)+c(a-c)\right]$
$=(a+b+c)\left[b c-a^{2}-b^{2}+a b+a c-c^{2}\right]$
$=(a+b+c)\left[-\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$
$=-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$
Hence, the determinant value is Negative
$\left(\right.$ Applying $\left.\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right)$
$=(a+b+c)\left|\begin{array}{lll}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{array}\right|$
[on taking $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\left.\mathrm{C}_{1}\right]$
$=(a+b+c)\left[1\left(b c-a^{2}\right)-b(b-a)+c(a-c)\right]$
$=(a+b+c)\left[b c-a^{2}-b^{2}+a b+a c-c^{2}\right]$
$=(a+b+c)\left[-\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$
$=-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$
Hence, the determinant value is Negative
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