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  • If x, y, z are different and <img alt="\begin{vmatrix} x &x^{2} &1+x^{3} \\ y& y^{2} &1+y^{3} \\ z& z^{2}&1+z^{3} \end{vmatrix}=0" src="https://learn.careers360.com/latex-image/?%5Cbegin%7Bvmatrix%7D%20x%20%26x%5E%7B2%7D%20%261&

If x, y, z are different 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 using properties of determinants show that 1 + xyz = 0.

 

 
 
 
 
 

Answers (1)

Given :

\begin{vmatrix} x &x^{2} &1+x^{3} \\ y& y^{2} &1+y^{3} \\ z& z^{2}&1+z^{3} \end{vmatrix}=0

\\ \left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|=0

\\ \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right|+x y z\left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|=0

(1+x y z)\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|=0

\\ \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \\ \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}

(1+x y z)\left|\begin{array}{ccc} x & x^{2} & 1 \\ y-x & y^2-x^2 & 0 \\ z-x & z^2-x^2 & 0 \end{array}\right|=0

(1+x y z)\left|\begin{array}{ccc} x & x^{2} & 1 \\ y-x & (y+x)(y-x) & 0 \\ z-x & (z+x)(z-x) & 0 \end{array}\right|=0

(1+x y z)(y-x)(z-x)\left|\begin{array}{ccc} x & x^{2} & 1 \\ 1 & y+x & 0 \\ 1 & z+x & 0 \end{array}\right|=0

\\ (1+x y z)(y-x)(z-x)1(z+x-(y+x))=0\\ (1+x y z)(y-x)(z-x)1(z-y)=0\\

As x,y,z are diffenernt.

Hence 1 + xyz = 0.

Posted by

Safeer PP

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