The equation obtained by eliminating a, b, c | vedclass

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(B) Given equations are:$x = \frac{a}{b-c} \Rightarrow a - bx + cx = 0$$y = \frac{b}{c-a} \Rightarrow ay + b - cy = 0$$z = \frac{c}{a-b} \Rightarrow a

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$\left|\begin{array}{ccc}1 & -x & x \ 1 & 1 & -y \ 1 & z & 1\end{array}ight|=0$

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(B) Given equations are:$x = \frac{a}{b-c} \Rightarrow a - bx + cx = 0$$y = \frac{b}{c-a} \Rightarrow ay + b - cy = 0$$z = \frac{c}{a-b} \Rightarrow az - bz - c = 0$These equations can be written in matrix form as:$\left|\begin{array}{ccc}1 & -x & x \ y & 1 & -y \ z & -z & -1\end{array}ight| \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$For a non-trivial solution of $a, b, c$,the determinant of the coefficient matrix must be zero:$\left|\begin{array}{ccc}1 & -x & x \ y & 1 & -y \ z & -z & -1\end{array}ight| = 0$Applying the column operation $C_1 ightarrow C_1 + xC_2 + xC_3$ or simply observing the structure,we can simplify the determinant. Alternatively,by expanding or manipulating rows/columns,we find that the determinant is equivalent to:$\left|\begin{array}{ccc}1 & -x & x \ 1 & 1 & -y \ 1 & z & 1\end{array}ight| = 0$Thus,option $(b)$ is correct.

The equation obtained by eliminating $a, b, c$ from the equations $x = \frac{a}{b-c}$,$y = \frac{b}{c-a}$,and $z = \frac{c}{a-b}$ is

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