KCET 2015 Math Solved Paper

Given equations,
x+y+z=6 → (1)
x+2y+3z=10 → (2)
x+2y+az=b→ (3)
Now [

1	1	1
1	2	3
1	2	a

]=[

6

10

b

]
R3→R3−R2, We have
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.