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I am a bit in doubt if I am doing this correct, I am row reducing and trying to get as close to a reduced echelon form, but is that even what I want? I am not sure if I am seeking the right answers with my solution

My solution

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You were doing fine, but the last step is just plain wrong. You already have the matrix in row echelon form after step 3; the last step messes you up. And your "translation" back to variables is very, very wrong. –  Arturo Magidin Jan 24 '12 at 21:46
    
Well I failed to say that it is a Argumented matrix, shouldnt I aim to get 1's in the right coloum then? –  Mech0z Jan 24 '12 at 22:18
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Not if the first column represents the first variable, etc. You aim to have a $1$ in the first nonzero entry of each row, and zeros under it, and any row of all zeros at the bottom, and for the first nonzero entry of each row to be further to the right than the first nonzero entry of each previous row, in order to get the row echelon form; for reduced row-echelon form, you also want the entries above those "leading 1s" to be zero. The matrix you had before the last step was already in row echelon form. –  Arturo Magidin Jan 24 '12 at 22:21

1 Answer 1

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First, there's no such thing as the solution to a matrix. What you are actually solving is a system of equations - in this case, a system of two equations in three unknowns - and you are using a matrix to represent the system of equations, and using matrix operations to solve the system.

I don't know what that 6 is that's sitting on top of your first matrix, so I propose to ignore it.

Your first matrix represents the system $$\eqalign{x_1+3x_2+4x_3&=7\cr 3x_1+9x_2+7x_3&=6\cr}$$

Your next-to-last matrix (which, as Arturo notes in the comments, is in row-echelon form) is $$\pmatrix{1&3&4&7\cr0&0&1&3\cr}$$ which represents the system $$\eqalign{x_1+3x_2+4x_3&=7\cr x_3&=3\cr}$$ From this, you can read off the value of $x_3$, and you can then get a formula for $x_1$ in terms of $x_2$; you can treat $x_2$ as a free parameter, able to take on any real value.

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How is it you determine a free variable? –  Mech0z Jan 25 '12 at 7:10
    
Sorry, I don't understand the question. If you have an equation like $a+3b=-5$, it doesn't have just one solution, it has infinitely many. You can let one of the two unknowns be anything you like - that makes it what I called a free parameter - and then you can work out what the other unknown has to be; you can solve for it, as a formula in terms of the free parameter. Try it! –  Gerry Myerson Jan 25 '12 at 12:36

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