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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $x,y,z$ are all distinct and $\begin{vmatrix}x&x^2&1+x^3\\y&y^2&1+y^3\\z&z^2&1+z^3\end{vmatrix}=0$ then the value of $\;\large\frac{-1}{xyz}$ is

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Given :
$\Rightarrow \begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+\begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix}=0$
$\Rightarrow \begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+xyz\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}=0$
$\Rightarrow (1+xyz)\begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}=0$
$\Rightarrow (1+xyz)(x-y)(y-z)(z-x)=0$
$(x\neq y\neq z)$
Hence $\large\frac{-1}{xyz}$$ = \large\frac{-1}{-1} $$ = 1$
answered Jan 30, 2014 by balaji.thirumalai
edited Jan 30, 2014 by balaji.thirumalai

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