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I'm doing some elementary work with matrices and I'm having a hard time distinguishing the operational differences between solving a linear system using augmented matrix method and from using Gauss Jordan elimination.

Correct me if I am wrong but to my understanding it seems like with augmented matrix method we have to find a form where it's in these types of form: $$\biggl[\begin{array}{cc|c} 1 & 0 & m \\ 0 & 1 & n \end{array}\biggr]$$ or $$\biggl[\begin{array}{cc|c} 1 & m & n \\ 0 & 0 & 0 \end{array}\biggr]$$ or… $$\biggl[\begin{array}{cc|c} 1 & m & n \\ 0 & 0 & p \end{array}\biggr]$$

while with Gauss elimination, our goal is to get the matrix into a reduced echelon form? It seems like the process for getting to those forms are the same am I correct? I am doing a practice test and one question asks me to solve the system using augmented while another question ask me to solve the system using Gauss elimination. Some clarity on this question would be appreciated.

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  • $\begingroup$ They are basically the same. The only difference is, when you transform the right hand side $\endgroup$ – user251257 Oct 10 '16 at 18:03
  • $\begingroup$ Care to elaborate? If not that's okay. Thanks though.t I will ask my professor tomorrow! $\endgroup$ – YangCPG Oct 10 '16 at 18:27
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There are basically three ways to make use of Gaussian elimination to solve a linear system. The first is to assemble the augmented matrix and put it into reduced row echelon form; from here you can read off the general solution to a particular linear system.

The second is to bring the matrix into echelon form, keeping track of how you did so, and in the process building an LU decomposition (or a $PA=LU$ decomposition, depending on how exactly you do it). You can then solve a linear system by two back-substitution steps.

The third is really a variant of the first, which is using Gaussian elimination to construct the inverse matrix (by augmenting with the identity matrix and converting to reduced row echelon form), and then solving a system by multiplying by the inverse matrix.

Both of the latter two methods can be used to start work on a system $Ax=b$ knowing only $A$ and not $b$. The former method requires you to already have $A$ and $b$ in order to start.

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  • $\begingroup$ Thank you so much. I know there was something that I'm not fully aware of yet! This explains it. $\endgroup$ – YangCPG Oct 11 '16 at 17:59

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