A) \[a+b+c\]
B) 0
C) \[{{b}^{3}}\]
D) \[ab+bc\]
Correct Answer: C
Solution :
\[{{A}^{-1}}={{A}^{2}}\]\[{{A}^{3}}=I\] Þ \[(a-6)({{b}^{2}}-ac)=0\Rightarrow {{b}^{2}}-ac=0\], \[=3\left[ \frac{-1+\sqrt{3}i}{2}-\frac{-1-\sqrt{3}i}{2} \right]=3\sqrt{3}\,i\] \[\therefore \] \[ac={{b}^{2}}\Rightarrow abc={{b}^{3}}.\]
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