Question

Given $a = \dfrac{x}{{y - z}}$ , $b = \dfrac{y}{{z - x}}$ and $c = \dfrac{z}{{x - y}}$ , where x , y , z are all non zero, find 1+ab+bc+ca.

Answer 1

Hint: Simplify all the three questions to find that the system of equations has infinite solutions. Use determinant equal to zero in those equations to find the required answer of the above question.

Complete step-by-step answer:

Simplifying all the three equations , we get
$a = \dfrac{x}{{y - z}} \Rightarrow ay - az = x \Rightarrow x - ay + az = 0$
$b = \dfrac{y}{{z - x}} \Rightarrow bz - bx = y \Rightarrow bx + y - bz = 0$
$c = \dfrac{z}{{x - y}} \Rightarrow cx - cy = z \Rightarrow - cx + cy + z = 0$

Thus we see that these systems of equations have infinite solutions . Thus their determinant = o .
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{ - a}&a \\
  b&1&{ - b} \\
  { - c}&c&1
\end{array}} \right| = 0\]
$ \Rightarrow \left( {1 + abc - abc} \right) - \left( { - ac - bc - ab} \right) = 0$
$ \Rightarrow 1 + ac + bc + ab = 0$

Note: Remember that the concept of the number of solutions of a given system of equations is used in such types of questions. Also recall how to solve determinants to get the accurate answer in these questions.