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UPSC NDA Math Model Paper 2

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Question : 65 of 120

Marks: +1, -0

If x, y, z are distinct real numbers and

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

xyz =

1
-1
2
-2

Solution:

Given, x, y, z are distinct real numbers and

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

To find the value of xyz, solve the determinant

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

= x[y^{2}(2+z^{3})-z^{2}(2-y^{3})] - x^{2}[y(2+z^{3})

-z(2+y^{3})] + (2+x^{3})y z^{2} - z y^{2}] = 0

= 2 x y^{2} + x y^{2} z^{3} - 2 x z^{2} + x y^{3} z^{2} - 2 x^{2} y

- x^{2} y z^{3} + 2 x^{2} z + x^{2} y^{3} z + 2 y z^{2} - 2 z y^{2}

+ x^{3} y z^{2} - x^{3} y^{2} z = 0

= 2 x y^{2} - 2 x z^{2} - 2 x^{2} y + 2 x^{2} z + 2 y z^{2}

- 2 z y^{2} + x^{3} y z^{2} - x^{3} y^{2} z + x y^{2} z^{3} + x y^{3} z^{2}

- x^{2} y z^{3} + x^{2} y^{3} z = 0

= 2(x y^{2} - x z^{2} - x^{2} y + x^{2} z + y z^{2} - z y^{2})

+ x y z(x y^{2} - x z^{2} - x^{2} y + x^{2} z

+ y z^{2} - z y^{2}) = 0

= (2 + x y z)(x y^{2} - x z^{2} - x^{2} y + x^{2} z

+ y z^{2} - z y^{2}) = 0

= (2 + xyz) = 0

= xyz = -2

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