System of linear equations with alpha variable

literacola
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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?
 
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literacola said:
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.
 
Cool, thanks! So what's the next step? Not sure where to go from here...
 
Can I get a hint?
 
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.
 
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