# Linear Algebra check please

1. Jan 17, 2009

### Sheneron

1. The problem statement, all variables and given/known data

Find values for a, b, and c such that the system has i) one solution ii) infinite solutions iii) no solutions.

$$x+5y+z=0$$
$$x+6y-z=0$$
$$2x+ay+bz=c$$

3. The attempt at a solution
I set it up like this:
$$x+5y+z=0$$
$$x+6y-z=0$$
$$0+(a-10)y+(b-2)z=c$$

So for one solution, I need to get it so that there is a single variable equal to a number. I chose a=0,b=2,c=-10. But I really could have chosen numerous different things as long as one variable was equal to a number.

For infinite I need to get it so that it eliminates the bottom equation to leave 2 equations and 3 unkowns, so I need a number equal to itself. I chose a=10,b=2,c=0. And that was the only choice I had.

For no solutions I need to get it so that there is a contradiction. I chose a=10,b=2, c=1. I could have chosen any number for c as long as it wasn't 0.

Are the answers and logic correct? Thanks in advance

2. Jan 17, 2009

### Staff: Mentor

Your logic seems fine, and your answers seem fine. You should be able to check for yourself, though, by putting the three sets of values back in the original system of equations to see that things work out as they should. Usually, the hard part is getting a solution, but checking it is pretty straightforward.

3. Jan 17, 2009

### Sheneron

Yes thank you, I guess I mainly wanted to make sure my thought process was sound.