Given equations, x+y+z=6 → (1) x+2y+3z=10 → (2) x+2y+az=b→ (3)
[
1
1
1
1
2
3
1
2
a
][
x
y
z
]=[
6
10
b
]
Now [
1
1
1
1
2
3
1
2
a
]=[
6
10
b
] R3→R3−R2, We have
[
1
1
1
1
2
3
0
0
a−3
]=[
6
10
b−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.