KCET2015MathematicsContinuity and Differentiability
The system of linear equations ( x+y+z=6, x+2 y+3 z=10 ) and ( x+2 y+a z=b ) has no solutions when
Options
- A( a=2, b 3 )
- B( a=3, b 10 )
- C( b=2, a=3 )
- D( b=3, a 10 )
Correct answer
B. ( a=3, b 10 )
Step-by-step solution
Given equations, x+y+z=6 (1)x+2 y+3 z=10 (2)x+2 y+a z=b (3) [ array llll 1 & 1 & 1 1 & 2 & 3 array ] [ array l x y 1 & 2 & a array ]= [ array c 6 10 array ] Now [ array lll 1 & 1 & 1 1 & 2 & 3 array ]= [ array c 6 10 array ]R₃ R₃-R₂ , We have [ array lll 1 & 1 & 1 1 & 2 & 3 0 & 0 & a-3 array ]= [ array c 10 b array ]=10 For now solutions a-3=0 =a=3 and b-10 0 b 10 Alternating Solution Comparing Eqs. ( 2 ) and (3), we have a=3, b=10 So, if a=3, b 10 then the system has no solution.