Matrices question (linear algebra)

calculusisrad
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Homework Statement


Give the general solution of the equation Ax=b in standard form.
The matrix is this: (sorry I can't do the long bracket like there should be)

[ 1 1 1 -1 0
2 0 4 1 -1
1 2 0 -2 2
0 1 -1 2 4] = A

[-1
10
-3
7] = b

Homework Equations



None


The Attempt at a Solution



Can you tell me if this is the correct reduced echelon form?

[ 1 1 -1 0
0 1 -1 -1 2
0 0 0 1 3
0 0 0 0 -7]

[-2
5
13
12]

I'm also having trouble finding the general solution, but I think it's because my reduced echelon form is wrong. I don't really even understand the entire concept of these types of problems so I'm struggling. I missed a couple days of class and am trying to figure it out from the book, but the answers aren't in the back so I can't check if I'm right... If you could tell me how to find the general solution, it would really help me. Thanks!
 
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calculusisrad said:

Homework Statement


Give the general solution of the equation Ax=b in standard form.
The matrix is this: (sorry I can't do the long bracket like there should be)

[ 1 1 1 -1 0
2 0 4 1 -1
1 2 0 -2 2
0 1 -1 2 4] = A

[-1
10
-3
7] = b

Homework Equations



None


The Attempt at a Solution



Can you tell me if this is the correct reduced echelon form?

[ 1 1 -1 0
0 1 -1 -1 2
0 0 0 1 3
0 0 0 0 -7]

[-2
5
13
12]

I'm also having trouble finding the general solution, but I think it's because my reduced echelon form is wrong. I don't really even understand the entire concept of these types of problems so I'm struggling. I missed a couple days of class and am trying to figure it out from the book, but the answers aren't in the back so I can't check if I'm right... If you could tell me how to find the general solution, it would really help me. Thanks!

When I submit your matrix A to Maple and ask for an LU decomposition, it gives me:
L = [[1 0 0 0],[2 1 0 0],[1 -1/2 1 0],[[0 -1/2 7 1]] (=[row 1, row 2, row 3, row 4]) and
U = [[1 1 1 -1 0],[0 -2 2 3 -1],[0 0 0 1/2 3/2],[0 0 0 0 -7]], which is not the same as your reduced row echelon form. I'm not sure exactly what you mean by a "reduced echelon form", because when I apply the Maple Command 'ReducedEchelonForm(A)' I get:
[[1 0 2 0 0],[0 1 -1 0 0],[0 0 0 1 0],[0 0 0 0 1]], which is not the same as U above, and not the same as your reduced echelon form.

I think you need to give us more explanations, and show us some of your work.

RGV
 
Well... That's not anything like my book says. According to my book, reduced echelon form is if every leading entry is one, and all the entries of the column above each leading entry are 0.

I can't really explain how it's done because its complicated...
 
calculusisrad said:
Well... That's not anything like my book says. According to my book, reduced echelon form is if every leading entry is one, and all the entries of the column above each leading entry are 0.

I can't really explain how it's done because its complicated...

Your will notice that "if every leading entry is one, and all the entries of the column above each leading entry are 0" from your book is exactly what the matrix L looks like from Maple. Maybe your book is using a different terminology from what I have seen elsewhere.

If you say that you can't explain how it's done because it's complicated, I don't see how anybody can help you. You certainly CAN show us the steps that YOU took in your attempt to solve the problem.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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