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AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 7 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, y, z

are all distinct, then

x y z =

-1
1
0
3

Solution:

\underset{(A)}{\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}} + \underset{(B)}{\begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix}} = 0

(given) ...(i)

A = \begin{bmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{bmatrix}

On applying

R_{3} \rightarrow R_{3} - R_{1}, R_{2} \rightarrow R_{2} - R_{1}

, we get

A = \begin{bmatrix} x & x^2 & 1 \\ y-x & y^2-x^2 & 0 \\ z-x & z^2-x^2 & 0 \end{bmatrix}

\therefore |A| = (y-x)(z^{2}-x^{2}) - (z-x)(y^{2}-x^{2})

= (y-x)(z-x)\{z+x-y-x\}

= (y-x)(z-x)(z-y)

B = \begin{bmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{bmatrix}

\Rightarrow |B| = \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix}

\Rightarrow |B| = xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}

R_{3} \rightarrow R_{3} - R_{1}, R_{2} \rightarrow R_{2} - R_{1}

\Rightarrow |B| = xyz \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix}

Then,

|B| = x y z \{(y-x)(z^{2}-x^{2}) - (z-x)(y^{2}-x^{2})\}

|B| = x y z (y-x)(z-x)(z-y)

From Eq. (i), we get

|A| + |B| = 0

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

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

\Rightarrow 1 + x y z = 0

[\because x, y, z]

are all distinct

\Rightarrow x y z = -1

⇒x y z =-1

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