 While Gaussian elimination is a perfectly good way of producing the row echelon form of a matrix, it almost always requires us to work with fractions. Fortunately, we can avoid using fractions using an ancient Chinese approach known as Fang Cheng Xu, which translates roughly as rectangular tabulation method. And this works as follows. Given whatever row we happen to be working on, we'll multiply all of the other rows by the pivot, and then we'll add multiples of the working row to these other rows. So let's see how this works. Let's go back to that awful, painful problem that got us those horrible fractions. But this time let's try and redo this problem using the Fang Cheng Xu method. So let's go ahead and set down our augmented coefficient matrix. Since we're already in standard form, that's just a matter of transcribing the coefficients in constants. The first row has pivot two, so we'll multiply all the other rows by two. So our first row remains unchanged. The second row was three, zero, two, five. So if we multiply it by two, we get six, zero, four, ten. And the third row, five, two, negative one, six, can be multiplied by two to get ten, four, negative two, twelve. And again, these new rows replace the original rows. So we don't need to keep the originals anymore. Next, I'll add multiples of the first row to the other so I can get zeros below the pivot. And because we multiplied every other row by the pivot, then the leading coefficient of every other row is going to be a multiple of the pivot. So if we take a look at the second row, we see that if we multiply the first row by negative three and add it, that will eliminate the leading coefficient. So we'll multiply the first row by negative three and add it to the second row, and that'll give us a new second row, zero, negative nine, one, negative eight. And since we have this new second row, we no longer need the two rows we added to produce it. Likewise, if we take a look at the third row, multiplying the first row by negative five and adding will eliminate the leading coefficient. So we'll multiply the first row by negative five and add it to get a new third row, zero, negative eleven, negative seven, negative eighteen. And again, now that we have a new third row, we don't need the two rows we use to produce it and we can eliminate them. So we'll write those down and we don't need the others. Since we have zeros below the first row pivot, we can move on to the second row where the leading coefficient is negative nine. So we'll multiply the following rows by negative nine. So that will give us zero, ninety nine, sixty three, and one hundred sixty two. And we no longer need the original third row, we can replace it with this new third row. Then we can add a multiple of the working row to get zeros below the pivot of the working row. In this case, if we multiply the working row by eleven, we get zero, negative ninety nine, eleven, negative eighty eight, and adding that to the third row gives us zero, zero, seventy four, seventy four. This gives us a new third row, which along with our first and second rows give us our augmented coefficient matrix in row echelon form. And now that the matrix is in row echelon form, we can use back substitution to solve for all the variables. So the last row corresponds to the equation seventy four z equals seventy four, which we solve z equals one. The next to last row, negative nine y plus z equals negative eight. We can substitute in our value for z equals one and solve for y. And finally the first row corresponds to the equation two x plus three y plus z equals six. We'll substitute in our values for y and z and solve the equation for x. While we won't be able to avoid fractions forever, we can delay their onset for quite some time as this example shows. So we'll start with our augmented coefficient matrix. The first row pivot is two, so we'll multiply the other two rows by two. So if we multiply the second row by two, we get six, two, four, two as our new second row. And if we multiply the third row by two, we get ten, six, zero, six as our new third row. Now to eliminate the entries below that first row pivot, we can multiply the first row by negative three and add it to the second row. Multiplying the first row by negative three gives us negative six, negative nine, negative three, negative twenty one. Then adding these entries to the second row gives us zero, negative seven, one, negative nineteen, which will be our new second row. Similarly, if we multiply our first row by negative five and add it to the last row, we'll get a new third row. So multiplying the first row by negative five gives us negative ten, negative fifteen, negative five, negative thirty five. And adding it to the third row gives us zero, negative nine, negative five, negative twenty nine, which will be our new third row. And this is the first step in our row reduction. Now we have zeros below the first row pivot, so we'll move on to the second row where the pivot is negative seven. So multiplying the following rows by negative seven, that gives us zero, sixty three, thirty five, two hundred and three as our new third row. Now if we multiply the second row by nine and add it to the third row, that will eliminate the entries below that second row pivot. So multiplying the second row by nine gives us zero, negative sixty three, nine, negative one hundred and seventy one. And adding it to the third row gives us zero, zero, forty four, thirty two, which will be our new third row. Now you'll notice one of the disadvantages here is that the numbers in our matrix grow larger and larger. And this is the price we have to pay for avoiding fractions. However, every now and then we find a row where all entries are divisible by the same number. And so it's convenient to remove any factor common to all terms in a row because this will help prevent our numbers from getting too large. So here we'll multiply that third row by one fourth to get zero, zero, eleven, eight as our new third row. And now our matrix is in row echelon form and we can solve using back substitution. So beginning with the last row, which corresponds to the equation eleven z equals eight, we can solve that z equals eight elevens. The next elast row corresponds to the equation minus seven y plus z equals negative nineteen. We know the value of z so we'll substitute that in and solve the equation for y. The first row corresponds to the equation two x plus three y plus z equal to seven. We'll substitute in our values for y and z and solve for x. And so this gives us our solution x equals negative twelve elevens, y equals thirty one elevens, and z equals eight elevens, which we'll write in vector form.