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KCET 2017 Math Solved Paper

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Question : 26 of 60

Marks: +1, -0

Let

\triangle = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix}

\triangle_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix}

then

$\triangle_1$ = -△
$\triangle_1$ = △
$\triangle_1$ ≠ △
$\triangle_1$ = 2△

Solution:

Given that,

\Delta = \begin{vmatrix} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{vmatrix}

→ (1)

and

\Delta_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ z y & z x & x y \end{vmatrix}

→ (2)

Take out common x, y and z in Eq.(1) from

R_1,R_2

and

R_3

respectively we get

\Delta = x y z \begin{vmatrix} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{vmatrix}

C_3 \rightarrow xyzC_3

,we get

\Delta = \begin{vmatrix} A & x & y z \\ B & y & z x \\ C & z & x y \end{vmatrix}

= \begin{vmatrix} A & B & C \\ x & y & z \\ y z & z x & x y \end{vmatrix} = \Delta_1

Therefore,

\triangle = \Delta_1

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