Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App