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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $x,y,z$ are all distinct and
$\left|
\begin{array}{ccc}
x & x^2 & 1+x^3 \\
y & y^2 & 1+y^3 \\
z & z^2 & 1+z^3
\end{array}
\right|=0$
then the value of $xyz$ is.
Solution
Expand the determinant by using column operation: $C_3 \rightarrow C_3 - C_1^3$ Then the determinant becomes a Vandermonde determinant multiplied by $(xyz+1)$. Since $x,y,z$ are distinct, the Vandermonde determinant is non-zero. Hence, $xyz+1=0$ $\Rightarrow xyz=-1$Online Test Series, Information About Examination,
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