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Here are two examples:

RowReduce[{{3, 1, a}, {2, 1, b}}]

evaluates to

{{1, 0, a - b}, {0, 1, -2 a + 3 b}}

but 

RowReduce[{{1, 2, 3, a}, {4, 5, 6, b}, {7, 8, 9, c}}]

evaluates to

{{1, 0, -1, 0}, {0, 1, 2, 0}, {0, 0, 0, 1}}

The result is independent of a, b and c.

Since I want to know the steps of reduction, I add a, b and c for bookkeeping. But it does not work in the second example. Is anything wrong or is any way to keep track of the steps of reduction?

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    $\begingroup$ Here's a hint: Solve[{x + 2 y + 3 z == a, 4 x + 5 y + 6 z == b, 7 x + 8 y + 9 z == c}, {x, y, z}] results in {}. $\endgroup$
    – JohnD
    Dec 13, 2012 at 18:30

2 Answers 2

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I'm not sure if your ultimate goal is a record of all of the row operations required to put a matrix in reduced row echelon form or if you want to figure out for what right-hand sides $\mathbf{b}=(a,b,c)$ does $A\mathbf{x}=\mathbf{b}$ have a solution.

For the latter,

RowReduce[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}]

which results in

{{1, 0, -1}, {0, 1, 2}, {0, 0, 0}}

Thus, $z$ is a free variable, $y=-2z$, and $x=z$. So only right-hand sides of the form $\mathbf{b}=\begin{bmatrix}1\\-2\\1\end{bmatrix}d$ where $d$ is any constant will work.

For the former, that is a different conversation altogether. The LU decomposition of $A$ deals with this.

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  • $\begingroup$ Thanks JohnD I think I try to solve something like x+2y+3z==Integers 4x+5y+6z==Integers 7x+8y+9z==Integers I do the Gaussian elimination without normalizing the leading coefficient in each row to 1 and get {{1,2,3},{0,-3,-6},{0,0,0}} So I will have x+2y+3z==Integers 3(y+2z)==Integers But RowReduce[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}] will result in {{1, 0, -1}, {0, 1, 2}, {0, 0, 0}} That is why I include {a,b,c} to keep track to the multiplicative constants. Do you know how to do rowreduce without normalizing each equation or ways to solve equations up to integers by mathematica? $\endgroup$
    – Charles
    Dec 13, 2012 at 19:53
  • $\begingroup$ So you want to solve equations where the right-hand side is constrained to integer values or when the unknowns (here, $x,y,z$) are constrained to integer values? $\endgroup$
    – JohnD
    Dec 13, 2012 at 20:27
  • $\begingroup$ The right-hand side is integer valued. No constraints on variables. $\endgroup$
    – Charles
    Dec 13, 2012 at 20:30
  • $\begingroup$ Well, from the answer above, $A\mathbf{x}=\mathbf{b}$ is only solvable for right-hand sides of the form $\mathbf{b}=(1,-2,1)d$, $d\in\mathbb{R}$. So if you only want integer right-hand sides, take $d\in\mathbb{Z}$. $\endgroup$
    – JohnD
    Dec 13, 2012 at 20:49
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    $\begingroup$ If you want to work over the integers use HermiteDecomposition instead of RowReduce. Your example: {uu, hh} = HermiteDecomposition[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}] {{{1, 0, 0}, {4, -1, 0}, {1, -2, 1}}, {{1, 2, 3}, {0, 3, 6}, {0, 0, 0}}} The first part of the result gives the transformation matrix. $\endgroup$
    – Daniel Lichtblau
    Dec 13, 2012 at 20:55
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I know this question is very old, but I stumbled across it while working on a similar problem and thought this might be of some use to someone in the future.

I think the OP intended the given matrix

{{1, 2, 3, a}, {4, 5, 6, b}, {7, 8, 9, c}}

to be taken as an augmented matrix with reduced row echelon form

{{1, 0, -1, 1/3 (-5 a + 2 b)}, {0, 1, 2, 1/3 (4 a - b)}, {0, 0, 0, a - 2 b + c}}

But

RowReduce[{{1, 2, 3, a}, {4, 5, 6, b}, {7, 8, 9, c}}]

knows nothing of the augmentation and continues the reduction by taking a - 2 b + c as the next pivot and gives

{{1, 0, -1, 0}, {0, 1, 2, 0}, {0, 0, 0, 1}}

A Workaround (Add a "Dummy" Column Vector)

A workaround that I came up with is to just add a "dummy" column vector Transpose[{{0, 0, 1}}] to the augmented matrix in the next-to-last column to absorb the pivot. Then

R = RowReduce[{{1, 2, 3, 0, a}, {4, 5, 6, 0, b}, {7, 8, 9, 1, c}}]

gives

{{1, 0, -1, 0, 1/3 (-5 a + 2 b)}, {0, 1, 2, 0, 1/3 (4 a - b)}, {0, 0, 0, 1, a - 2 b + c}}

You can then either simply ignore the dummy column in the result or, if desired, do

Drop[R, None, {-2}]

which gives

{{1, 0, -1, 1/3 (-5 a + 2 b)}, {0, 1, 2, 1/3 (4 a - b)}, {0, 0, 0, a - 2 b + c}}
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