If system of equations cx+y+1=0 x+cy+2=0 x+y+1=0 is consistent then the values of c is equal to
A
2
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B
3
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C
4
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D
5
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Solution
The correct option is B2 D=∣∣
∣∣c111c2111∣∣
∣∣=0 ⇒c2−3c+2=0⇒c=1,2 for c=1 system become x+y+1=0 x+y+2=0 x+y+1=0 which is inconsistent system Now for c=2 2x+y+1=0 x+2y+1=0 x+y+1=0 which are concurrent lines ⇒c=2