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