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EAMCET Engineering 2010 Solved Paper

Section: Mathematics

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Question : 66 of 160

Marks: +1, -0

If

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

1+xyz

0
-1
1
2

Solution:

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

\; \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

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

C_1 \leftrightarrow C_2

and

C_2 \leftrightarrow C_3

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

\; (1+xyz) \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = 0

\because R_2 \rightarrow R_2 - R_1

R_3 \rightarrow R_3 - R_1

(1+xyz) \begin{vmatrix} x & x^2 & 1 \\ y-x & y^2-x^2 & 0 \\ z-x & z^2-x^2 & 0 \end{vmatrix} = 0

Expand with

r

to

C_3

\; (1+xyz)

\; \{ (y-x)(z-x)(z+x) - (z-x)(y-x)

\; \;\;\; (y+x) \} = 0

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

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

\; \because \;\;\; x \neq y \neq z \Rightarrow xyz+1 = 0

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