Determinants Matrices Question 91
Question: Given $ a=x/(y-z), $
$ b=y/(z-x) $ and $ c=z/(x-y), $ where x, y and z are not all zero, Then the value of $ ab+bc+ca $
Options:
A) $ 0 $
B) $ 1 $
C) $ -1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ a=x/(y-z)\Rightarrow x-ay+az=0…(1) $
$ b=y/(z-x)\Rightarrow bx+y-bz=0…(2) $
$ c=z/(x-y)\Rightarrow -cx+cy+z=0…(3) $ As x, y, z are not all zero, the above system has a non-trivial solution so $ \Delta = \begin{vmatrix} 1 & -a & a \\ b & -1 & -b \\ -c & c & 1 \\ \end{vmatrix} $
$ \therefore 1+ab+bc+ca=0 $
BETA
