# Math Help - gaussian elimination solutions

1. ## gaussian elimination solutions

I have used gaussian elimination to solve a system of 3 equations in x, y and z. I have the start of the answer but I'm not sure why the answer says a is not equal to 3. I can see a= -1 would cause an undefined result. Here is a link to the scan: ImageShack&#174; - Online Photo and Video Hosting

I have not been able to upload nor to upload from a website.

While getting to the solution for z there was factors of (a-3) that cancelled out.

2. What is the original problem?

3. The question involves large amounts of algebra that I did outside of the matrices. The equations are x+2y-z=b, ax+3y+z=b, 2x+y+(a-1)z=0. The question says "Find the solutions discussing the various cases that arise for different values of the constants a, b."

4. Interesting. Ok, so the determinant of the coefficient matrix is

$-2a^{2}+4a+6=-2(a-3)(a+1).$

So, if a = 3 or a = -1, you have the possibility of infinitely many solutions, depending on what b is. Apparently, if a = -1, then you must have b = 0, or the system is inconsistent. a = 3 is a possibility, but you'll get infinitely many solutions.

To see what's really going on, I might recommend solving the system three separate times: a = -1, a = 3, and a something else.

5. I had started doing that. I know how to calculate the determinant. I don't think I've used it for anything more than finding an inverse matrix. What is it saying?

6. If the determinant is nonzero, the matrix is invertible, which means you'd get one unique solution, period. That is, you start with the matrix equation

$Ax=b,$

and since A is invertible, you can left-multiply by A inverse to obtain

$A^{-1}Ax=x=A^{-1}b.$

So, I've just shown that there is a solution, and it's $x=A^{-1}b.$ Now, still assuming invertibility, let's prove uniqueness. Suppose I have two solutions of the system, x and y. Then

$Ax=b$ and $Ay=b.$

Hence,

$Ax=Ay.$

But, since A is invertible, I can multiply both sides of this equation on the left by A inverse to obtain

$A^{-1}Ax=A^{-1}Ay,$ or

$x=y.$

Hence, invertibility guarantees us one unique solution. And a nonzero determinant guarantees us an invertible coefficient matrix A. So a nonzero determinant guarantees us that the system is consistent (thus, there is at least one solution), and that there is only one solution (that is, there is at most one solution).

However, if the determinant is zero, then the matrix is not invertible. You could have infinitely many solutions, or you could have no solutions, depending on the vector b.

Does that all make sense?

7. Yes, thanks. Is this saying roughly the same thing?
ImageShack&#174; - Online Photo and Video Hosting

8. I think my post had a little more detail about invertibility and the determinant, but your linked pic there certainly isn't saying anything different from what I was saying.