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I was given the matrix: $$\left[\begin{matrix}1&3&0&-2&-7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$$ And I was told to continue the row operations to solve the solution for the original system. So for the first step I subtracted $3R_4$, where $R_4$ is row $4$, to get rid of the $x_2$ in the original system. To get: $$\left[\begin{matrix}1&3&0&-2&-7\\0&1&0&0&12\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$$ Then what I did was $R_1-3R_2$ to get rid of the $x_2$:

$$\left[\begin{matrix}1&0&0&-2&-43\\0&1&0&0&12\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$$Then lastly $R_1+2R_4$ to get rid of the $x_4$ in $R_1$

$$\left[\begin{matrix}1&0&0&0&-47\\0&1&0&0&12\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]$$I'm curious as to why were able to do this to a matrix. It seems like these manipulations to the rows would make it different. It's strange to me that we can add a constant times another row to get a new, replacement row. Is there any theory behind this, or is it just that simple that we can just change it?

Edit: I changed the question from "did I do this right to" a "why can we do this" just so theres no confusion about the current comments.

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Looks fine to me... – DonAntonio Jan 27 '13 at 20:50
Correct according to Mathematica – valtron Jan 27 '13 at 20:51
just as a matter of nomenclature, you reduced the matrix correctly. you need some other matrices/equations to be solving things. – MSEoris Jan 27 '13 at 20:54
Thanks for the confirmation, I just changed the question a little bit. I meant to make this a two part question but forgot about the second part. – TheHopefulActuary Jan 27 '13 at 20:56
up vote 3 down vote accepted

We can add a constant multiple ($k$) of a row to another without changing the system's solutions because $A=B$ and $C=D$ if and only if $A+kC=B+kD$ and $C=D$.

Similarly, we can multiply a row by $k\neq 0$ without changing the solutions because $A=B$ if and only if $kA=kB$.

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Yes, you are right.It is correct :)

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thanks for the confirmation, I changed the question a little, if you know how to answer could you? Thanks! – TheHopefulActuary Jan 27 '13 at 20:57
have a look at here :… – tevlon84 Jan 27 '13 at 21:03

Answer to the 'why' it works part of the question. There are three kinds of what are called elementary row operations on an augmented matrix to solve a system of linear equations of this form.

They are

  1. Interchange two rows. (not needed for your question). To undo this, in the new matrix just interchange the same two rows to get back to the original matrix.

  2. Multiply a row by a nonzero constant. (not needed for your question). To undo this, in the new matrix divide the same row by that constant to get back to the original matrix.

  3. Add a multiple of one row,say row i, from a different row, say row j. (you did this correctly twice). To undo this, subtract the same multiple of the new row i from the new row j.

Since you can undo each of these operations and the rows represent linear equations, all of the matrices you produce will correspond to linear systems of equations with the same solutions as the original system.

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Each of these elementary row operations can be applied to the identity matrix to get a matrix which when multiplying on the left of the matrix of the original system will give you that elementary row operation applied to the original matrix. – Barbara Osofsky Jan 28 '13 at 0:44

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