I wrote this in September so that I could do matrix row reduction on my TI-89. It works, usually, but sometimes it doesn't row reduce all the way. Any ideas what I've done wrong? Or is it just some strange quirk of the calculator?(adsbygoogle = window.adsbygoogle || []).push({});

If you don't know TI-89 code, it is self explanatory except for a few built in matrix functions which I explained with C++ style comments.

The matrix I am trying to do isCode (Text):

solvemat(x) // the name of the function. x is the matrix

Func //stating that it is a function and not a program

Local j, L, i, k, o // These are integers. Note:

//Sorry about the non-

//descriptive names, I had to type on an

//alphabetic non-qwerty

//keyboard so I didn't want them to be long

// j and L will be dealt with shortly

// i is the loop variable for columns

// k is the loop variable for rows

// o keeps track of the last row that properly starts with a 1

colDim(x) - 1 -> j //colDim(x) = number of columns in x. -> j

// means "assign to j"

rowDim(x) -> L //rowDim(x) = number of rows in x

0->o //o is initialized to 0

For i, 1, j, 1 // for(i = 1; i <= j; i++)

o+1 -> k

While x[k, i] = 0

k + 1 -> k

If k > L

Goto down //I know it's lame but there

EndWhile //is no "continue" in TI-89 basic

mRow(1/(x[k, i]), x, k) -> x

/* this built-in matrix solving function means

"multiply row k of matrix x

by 1/(x[k, i]) and put the result back into

matrix x." What I am doing here is putting

a 1 at the beginning of the row */

rowSwap(x, k, o+1) // swaps row k and row o+1

o+1 -> o //I could have done this a line earlier

// and had neater code, but I didn't

/* Now I have a row that is in the proper location, and it has

a 1 at the beginning of it. Now I will use that row to

eliminate all the nonzero elements of the matrix above and below the 1. */

For k, 1, L, 1 // for(k = 1; k <= L; k++) also L =

// rowDim(x) from earlier

If k = o // that's an 'o'

Goto d2

If x[k, i] != 0

mRowAdd(-x[k, i], x, o, k) -> x

/* This built-in matrix solving function means

"multiply row o of matrix x by -x[k, i] and add the

result to row k, and the resulting matrix goes back

into matrix x */

Lbl d2

EndFor

Lbl down

EndFor

Return x

EndFunc

1 -1 0 5

-1 1 5 2

0 1 1 0

The solution of this matrix is

1 0 0 18/5

0 1 0 -7/5

0 0 1 7/5

But my function, solvemat(x), spits this out instead

1 1/4 0 13/4

0 5/4 0 -7/4

0 -1/4 1 7/4

If I do solvemat(solvemat(x)) it finishes the row reduction and returns the correct result. And the strangest thing is that if I alter the "1" in position (2, 2) of the original matrix x, the program runs correctly even if I only altered the 1 to, say, 1.000001.

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# Homework Help: Row Reduction Program

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