The equation obtained by eliminating \(a, b, c\) from the equations \(x=\frac{a}{b-c}, y=\frac{b}{c-a}\),…
The equation obtained by eliminating \(a, b, c\) from the equations \(x=\frac{a}{b-c}, y=\frac{b}{c-a}\), \(z=\frac{c}{a-b}\) is
- \(\left|\begin{array}{lll}1 & -x & x \\ 1 & -y & y \\ 1 & -z & z\end{array}\right|=0\)
- \(\left|\begin{array}{ccc}1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1\end{array}\right|=0\)
- \(\left|\begin{array}{ccc}1 & -x & x \\ y & 1 & -y \\ -z & z & -1\end{array}\right|=0\)
- \(\left|\begin{array}{lll}x & y & 1 \\ y & x & 1 \\ 1 & x & y\end{array}\right|=0\)
Solution
Given equations
\(\begin{gathered}
x=\frac{a}{b-c} \Rightarrow a-b x+c x=0 \\
y=\frac{b}{c-a} \Rightarrow a y+b-c y=0
\end{gathered}\)
and \(\quad z=\frac{c}{a-b} \Rightarrow a z-b z-c=0\)
Now, on eliminating \(a, b, c\) from the above equations, we get
\(\left|\begin{array}{ccc}
1 & -x & x \\
y & 1 & -y \\
z & -z & -1
\end{array}\right|=0\)
On applying \(C_1 \rightarrow C_1+C_2+C_3\), we get
\(\left|\begin{array}{ccc}
1 & -x & x \\
1 & 1 & -y \\
-1 & -z & -1
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
1 & -x & x \\
1 & 1 & -y \\
1 & z & 1
\end{array}\right|=0\)
Hence, option (b) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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