A) \[x=0\]
B) \[x=a\]
C) \[x=b\]
D) \[x=c\]
Correct Answer: A
Solution :
Obviously, on putting \[x=0\], we observe that the determinant becomes \[{{\Delta }_{x=0}}=\left| \,\begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix}\, \right|=a(bc)-b(ac)=0\] \[\therefore \] \[x=0\] is a root of the given equation. Aliter : Expanding \[\Delta \], we get \[\Delta \equiv -(x-a)\,[-(x+b)(x-c)]+(x-b)\,[(x+a)\,(x+c)]=0\] \[\Rightarrow \] \[2{{x}^{3}}-(2\,\Sigma ab)x=0\] \[\Rightarrow \] Either \[x=0\] or \[{{x}^{2}}=\sum{ab}\] (i.e., \[x=\pm \sum{ab})\] Again \[x=0\] satisfies the given equation.You need to login to perform this action.
You will be redirected in
3 sec