Question
Given $$a = \frac{x}{{\left( {y - z} \right)}},b = \frac{y}{{\left( {z - x} \right)}}$$ and $$c = \frac{z}{{\left( {x - y} \right)}},$$ where $$x, y$$ and $$z$$ are not all zero, Then the value of $$ab + bc + ca$$
A.
$$0$$
B.
$$1$$
C.
$$ - 1$$
D.
None of these
Answer :
$$ - 1$$
Solution :
$$\eqalign{
& a = \frac{x}{{\left( {y - z} \right)}} \cr
& \Rightarrow \,x - ay + az = 0\,\,\,.....\left( 1 \right) \cr
& b = \frac{y}{{\left( {z - x} \right)}} \cr
& \Rightarrow \,bx + y - bz = 0\,\,\,.....\left( 2 \right) \cr
& c = \frac{z}{{\left( {x - y} \right)}} \cr
& \Rightarrow \, - cx + cy + z = 0\,\,\,.....\left( 3 \right) \cr} $$
As $$x, y, z$$ are not all zero, the above system has a non-trivial
solution so, \[\Delta = \left| {\begin{array}{*{20}{c}}
1&{ - a}&a\\
b&{ - 1}&{ - b}\\
{ - c}&c&1
\end{array}} \right|\]
∴ $$1 + ab + bc + ca = 0$$