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Mathematics Question on Determinants

If x,yx, y and zz are all different and ∣xx21+x3 yy21+y3 zz21+z3∣=0\begin{vmatrix}x&x^{2}&1+x^{3}\\\ y&y^{2}&1+y^{3}\\\ z&z^{2}&1+z^{3}\end{vmatrix}=0 then

A

xyz=−1xyz = -1

B

xyz=1xyz = 1

C

xyz=−2xyz = -2

D

xyz=2xyz = 2

Answer

xyz=−1xyz = -1

Explanation

Solution

The correct option is(A): xyz =−1.

Given , ∣xx21+x3 yy21+y3 zz21+z3∣=0\left|\begin{matrix}x&x^{2}&1+x^{3}\\\ y&y^{2}&1+y^{3}\\\ z&z^{2}&1+z^{3}\end{matrix}\right|=0
⇒∣xx21 yy21 zz21∣+∣xx2x3 yy2y3 zz2z3∣=0\Rightarrow \left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|+\left|\begin{matrix}x&x^{2}&x^{3}\\\ y&y^{2}&y^{3}\\\ z&z^{2}&z^{3}\end{matrix}\right|=0
⇒∣xx21 yy21 zz21∣+xyz∣1xx2 1yy2 1zz2∣=0\Rightarrow \left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|+xyz \left|\begin{matrix}1&x&x^{2}\\\ 1&y&y^{2}\\\ 1&z&z^{2}\end{matrix}\right|=0
⇒∣xx21 yy21 zz21∣+xyz∣xx21 yy21 zz21∣=0\Rightarrow \left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|+xyz\left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|=0
⇒(1+xyz)∣xx21 yy21 zz21∣=0\Rightarrow \left(1+xyz\right) \left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|=0
⇒(1+xyz)∣xx21 yy21 zz21∣=0\Rightarrow\left(1+xyz\right)\left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|=0
⇒1+xyz=0\Rightarrow 1+xyz=0
and ∣xx21 yy21 zz21∣≠0\left|\begin{matrix}x&x^{2}&1\\\ y&y^{2}&1\\\ z&z^{2}&1\end{matrix}\right|\ne0
∴xyz=−1\therefore xyz=-1