Solving for Nontrivial Solutions of a System of Linear Equations

[nietzsche]

Homework Statement

For what values of a does the system

(a-1)x + 2y = 0
2x + (a-1)y = 0

have a nontrivial solution?

Homework Equations

The Attempt at a Solution

Argh... I'm really bad at linear algebra, can't seem to grasp the concepts.

I know that if I make this into a matrix

<br /> \left[<br /> \begin{array}{cc}<br /> a-1 &amp; 2\\<br /> 2 &amp; a-1<br /> \end{array}<br /> \right]<br />

that the only way the system will have a nontrivial solution is if the reduced row echelon form is not the 2x2 identity, i.e. the matrix is singular. But I don't know how to use those ideas to solve it. Any help would be great!

 

[Pengwuino]

If the determinant is 0, what does that tell you about the system?

 

[nietzsche]

If the determinant is 0, what does that tell you about the system?

For some reason, my prof hasn't taught us anything about the determinant. I've read about it, and I sort of have an idea of how it works, but I don't think our prof wants us to use it to solve problems.

 

[Dick]

Ok, so start reducing it to echelon form. Divide the first row by a-1. Under what conditions can you fail to get ones on the diagonal?

 

[nietzsche]

my intuition is telling me that a = 3 ... i did it like this:

<br /> \left[<br /> \begin{array}{cc}<br /> 2 &amp; a-1\\<br /> a-1 &amp; 2<br /> \end{array}<br /> \right]<br /> \\<br /> \text{then}<br /> \\<br /> \left[<br /> \begin{array}{cc}<br /> 1 &amp; \frac{a-1}{2}\\<br /> \frac{a-1}{2} &amp; 1<br /> \end{array}<br /> \right]<br />

so from this we see that \frac{a-1}{2} = 1 because that would make row 2 a multiple of row 1, and subtracting it out would leave a row of zeros. so a = 3.

does it make sense? i feel like I'm missing something.

 

[Dick]

That makes a lot of sense. Sure a=3 is a problem. But keep going with the row reduction. Multiply the first row by (a-1)/2 and subtract it from the second row. There is another value of a that creates a problem.

 

[nietzsche]

i was thinking that maybe a = 1 is also a problem, but if you substitute a = 1 into the original matrix, you can transform it into the 2x2 identity.

but if i try to put it into rref i end up having to divide by a-1, and if a=1 then it is undefined.

so I'm confused about that...

 

[Dick]

my intuition is telling me that a = 3 ... i did it like this:

<br /> \left[<br /> \begin{array}{cc}<br /> 2 &amp; a-1\\<br /> a-1 &amp; 2<br /> \end{array}<br /> \right]<br /> \\<br /> \text{then}<br /> \\<br /> \left[<br /> \begin{array}{cc}<br /> 1 &amp; \frac{a-1}{2}\\<br /> \frac{a-1}{2} &amp; 1<br /> \end{array}<br /> \right]<br />

so from this we see that \frac{a-1}{2} = 1 because that would make row 2 a multiple of row 1, and subtracting it out would leave a row of zeros. so a = 3.

does it make sense? i feel like I'm missing something.

Like I edited my last post to say, just multiply the first row by (a-1)/2 and subtract from the second row. Then ask yourself when you can get a second row of zeros.

 

[nietzsche]

ah, i see

a = -1

thanks again dick. you're really saving my skin.

 

[Dick]

ah, i see

a = -1

thanks again dick. you're really saving my skin.

Well, you are helping by using the hints and thinking about them. Not everyone does that.