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? 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. Code (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 The matrix I am trying to do is 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.