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If x + y + z = 0, prove that  \left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|

Answers (1)

Given LHS,

\left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|

Expand along Row 1

\\\\ = xa$ \{ $ (za)(ya) -(xc)(xb)$ \} $ -(yb)$ \{ $ (yc)(ya) -(zb)(xb)$ \} $ + (zc)$ \{ $ (yc)(xc) -(zb)(za)$ \} $ \\ \\ \vspace{\baselineskip}= xa$ \{ $ a\textsuperscript{2}yz -x\textsuperscript{2}bc$ \} $ -yb$ \{ $ y\textsuperscript{2}ac -b\textsuperscript{2}xz$ \} $ + zc$ \{ $ c\textsuperscript{2}xy -z\textsuperscript{2}ab$ \} $ \\ \\ \vspace{\baselineskip}= a\textsuperscript{3}xyz -x\textsuperscript{3}abc -y\textsuperscript{3}abc + b\textsuperscript{3}xyz + c\textsuperscript{3}xyz -z\textsuperscript{3}abc\\ \\ \vspace{\baselineskip}= xyz(a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3}) -abc(x\textsuperscript{3} + y\textsuperscript{3} + z\textsuperscript{3})\\

 

Given x+y+z = 0

\vspace{\baselineskip} \Rightarrow x^3 + y^3 + z^3 = 3xyz\\ \vspace{\baselineskip} = xyz(a^3 + b^3 + c^3) - abc(3xyz)\\ \vspace{\baselineskip} = xyz(a^3 + b^3 + c^3 - 3abc)\\

=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|

 

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