Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions

Question

Hi, i reduced this however i am stuck when it comes to finding the 3 solutions, (infinity, one , no solutions). I am confused with what do they mean with them! would appreciate it if someone can answer what do they mean in general and in this question!

The matrix after reduction:

1 1 0 2

0 1 1 2

0 0 2 2

0 0 (a-b+c) -2a

Question about my solution: Here can we say that z=1

now solving for last row:

(a-b+c)z=-2b -.a-b+c = -2b -> a+c=-b ->b=a+c

Now:

a+b+c = 0 ->a+c=-b -> a+c = -(a + c) so 0=0 Thus: infinitely many solutions when a+b+c = 0

and unique solution when a+b+c (not =) 0?

Answer 1

I'm basing this answer strictly on the matrix you gave, since the equations you wrote below that don't seem to agree with the matrix.

Your matrix indicates the system has either no solution or exactly one solution (x=1,y=1,z=1).

If you continue row reducing, you can write the last row as

0 0 0 (b-3a-c). So if b-3a-c=0, you get a row of all zeros. However, that would mean there's only one solution, not infinitely many. This is because there are only 3 unknowns, so the first three rows give the unique solution. On the other hand, if b-3a-c is not zero, then the last row reduces to 0 0 0 1, which implies the system has no solution.