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I am wondering if there is a Mathematica command that will put a matrix in row echelon form. That is, put

$$ \begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 4\\ -1 & 0 & 2 \end{bmatrix} $$ in row echelon form: $$ \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix} $$ I am aware that I can do a sequence of elementary row operations. I am also aware of the RowReduce command which puts a matrix in reduced row echelon form. I even saw a Method->"OneStepRowReduction" used by the RowReduce command which I thought might be the choice.

Just wondering if there is a command for this that I cannot find.

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    $\begingroup$ Your second matrix is not in row echelon form. $\endgroup$ – enzotib Mar 22 '15 at 13:21
  • $\begingroup$ What @enzotib quite correctly observes leads me to wonder if perhaps you actually want the "echelon" form over the ring of integers. That would be found with HermiteDecomposition. $\endgroup$ – Daniel Lichtblau Mar 22 '15 at 13:27
  • $\begingroup$ Fixed the row echelon error. Sorry about that. No, I just want the ordinary row echelon form students have to do when first introduced to Gaussian elimination. $\endgroup$ – David Mar 22 '15 at 15:12
  • $\begingroup$ Note that there is nothing unique about a row echelon form, even such a form whose leading entries are all 1. So the question needs to be restated in a more precise manner. E.g., do you want a command that follows a particular algorithm for obtaining a row echelon form? As you noted, it's easy to do a sequence of elementary row operations, and to write little functions to do such operations, then to write a main function that calls those little functions to follow a particular reduction algorithm. But then how do you want to handle roundoff errors? (Or do you want exact rational arithmetic?) $\endgroup$ – murray Mar 22 '15 at 20:24
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I think LUDecomposition should be able to do this. Using code from help in Mathematica docs:

a = {{1, 2, 3}, {2, 3, 4}, {-1, 0, 2}};
{lu, p, c} = LUDecomposition[a]
(u = lu SparseArray[{i_, j_} /; j >= i -> 1, {3, 3}]) // MatrixForm

Mathematica graphics

If you want all pivots to be +1, this can now be easily done.

d = Position[Diagonal[u], -1]
(u[[First@#]] = -u[[First@#]]) & /@ d

Mathematica graphics

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