question_answer

The value of  is:

A) \[abc\]             

B) \[a+b+c\]      

C) 4abc                            

D) 0

Correct Answer: C

Solution :

[c]  \[{{R}_{1}}\to {{R}_{1}}\to ({{R}_{2}}+{{R}_{3}})\] \[{{R}_{2}}\to c.{{R}_{2}}-b.{{R}_{3}}\] Take c common from \[{{R}_{3}}\] = \[=\frac{1}{c}\times c.\text{ }\left( -\text{ }2bc \right)\left[ c\text{ }-a-b-c-a+b \right]\] \[=\left( -2bc \right)\left( -2a \right)=4abc\] Hence, option [c] is correct.