System of linear equations with alpha variable

[literacola]

Homework Statement

For which value of alpha does the following system of linear equations have a solution?

x + y = 2
x +αy = -1
αx + y = 1

The Attempt at a Solution

I put it into a matrix that looks like this:

1 1 | 2
1 α | -1
α 1 | 1

Then I subtracted row1 from row2 and it made the matrix look like:

1 1 | 2
0 (α-1) | -1
α 1 | 1

Then I took the 2nd row as α -1 = -1 and solved α to be zero. So, does this mean that the solution for this is all numbers α such that α is not equal to zero?

 

[gabbagabbahey]

I put it into a matrix that looks like this:

1 1 | 2
1 α | -1
α 1 | 1

Then I subtracted row1 from row2 and it made the matrix look like:

1 1 | 2
0 (α-1) | -1
α 1 | 1

Then I took the 2nd row as α -1 = -1 and solved α to be zero.

First, when you subtract the first row from the second you get

\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & \alpha-1 & -3 \\ \alpha & 1 & 1 \end{array}\right]Second, your second row tell you (\alpha-1)y=-3, not (\alpha-1)=-3

So, does this mean that the solution for this is all numbers α such that α is not equal to zero?

No.

 

[literacola]

Cool, thanks! So what's the next step? Not sure where to go from here...

 

[literacola]

Can I get a hint?

 

[gabbagabbahey]

I think the easiest method is to simply say that if (\alpha-1)y=-3, then y=\frac{3}{1-\alpha} and substitute that into any of your 3 equations, solve for x and sub those into the remaining equations to get an equation for \alpha.