 One problem with Gaussian elimination is that we almost always encounter fractions. So let's start out with this system of equations, which is already written in standard form, which means we can put down our augmented coefficient matrix almost immediately. Now our first row pivot is 2, so we need to multiply that first row by 1 half to make that pivot equal to 1. And so we'll multiply each term in that first row by 1 half. 2 times a half is 1, 3 times a half is 3 halves, 1 times a half is 1 half, and 6 times a half is 3. We no longer need the original first row and the other two rows remain unchanged. Next, if we multiply this first row by negative 3 and add it to the second row, that will eliminate the entry below the first row pivot. So we'll keep our first row and we'll multiply each term by negative 3. So 1 times negative 3 is negative 3, 3 halves times negative 3 is negative 9 halves, 1 half times negative 3 is negative 3 halves, and 3 times negative 3 is negative 9. Now we'll add this row to the second row and we'll copy down that second row closer for convenience. So negative 3 plus 3 is 0, negative 9 halves plus 0 is negative 9 halves, negative 3 halves plus 2 is 1 half, and negative 9 plus 5 is negative 4. Now we can also record this as minus 3 r1 plus r2 replaces r2. And we'll keep the original first row, the new second row, and the original third row. And the other rows are unnecessary, so we don't have to keep track of them any. Now we'll want to eliminate the leading entry in the third row. So if we multiply the first row by negative 5 and add it to the last row, we'll get a 0 there. So negative 5 times 1 gives us negative 5, negative 5 times 3 halves gives us negative 15 halves, negative 5 times 1 half gives us negative 5 halves, and negative 5 times 3 gives us negative 15. We'll copy down the original third row closer for convenience and we'll add negative 5 plus 5 is 0, negative 15 halves plus 2 is negative 11 halves, negative 5 halves minus 1 is negative 7 halves, and negative 15 plus 6 is negative 9. So our new coefficient matrix is going to include the original first row, which we didn't change. The second row, which we also left unchanged, and our new third row. And we'll go ahead and copy down our new augmented coefficient matrix for reference. So our first row 1, 3 halves, 1 half, 3. Our second row 0, negative 9 halves, 1 half, negative 4. And our third row 0, negative 11 halves, negative 7 halves, negative 9. Now we'll continue the Gaussian elimination by moving on to the second row. Since the leading coefficient in the second row is negative 9 halves, we'll multiply everything in that second row by negative 2 ninths to make that leading coefficient equal to 1. So our first row will leave unchanged and our second row will copy down close by for reference. 0 times negative 2 ninths is still 0, negative 2 ninths times negative 9 halves is 1, negative 2 ninths times 1 half is negative 1 ninths, and negative 2 ninths times negative 4 is 8 ninths. And the result will be our new second row, so we can erase the old row. And since we haven't done anything with our third row, we'll copy that over unchanged. Next, since the leading entry in our third row is negative 11 halves, if we multiply our new second row by 11 halves and add it to the third row, we'll eliminate that leading entry. So the first row will be unchanged, the second row will be our new second row. We'll multiply that by 11 halves to get 0, 11 halves, negative 11 eighteenths, and 88 eighteenths. Adding this to our third row, 0 plus 0 is 0, 11 halves plus negative 11 halves is 0, negative 11 eighteenths minus 7 halves is negative 74 eighteenths, 88 eighteenths minus 9 is negative 74 eighteenths. Finally, we'll make the leading coefficient in the third row equal to 1 by multiplying it by negative 18 74th. So multiplying that third row by negative 18 74th is going to make it 0, 0, 1, 1. And this gives us our final row reduced form, our first row 1, 3 halves, 1 half, 3, our second row 0, 1, negative 1 ninths, 8 ninths, and our third row 0, 0, 1, 1. And now that we're in row echelon form, we can solve the system through back substitution. So our last row corresponds to the equation z equals 1. Our next to last row corresponds to the equation y minus 1 ninths, z equals 8 ninths. And the first row corresponds to the equation x plus 3 halves y plus 1 half z is equal to 3. And now we can try solving our system of equation through back substitution. Starting with the last equation z equals 1. We'll solve that for z. There it is. Our next to last equation y minus 1 ninths z equals 8 ninths. We'll solve that for y, substituting our value of z, and find that y is equal to 1. And our first equation, we'll solve that for x, substituting y and z, and solve for x. And this gives us our solution x equals 1, y equals 1, z equals 1, or as a vector 1, 1, 1. And we notice that even though our system of equations has coefficients and constants all equal to integers, and our solutions are also all equal to integers, the process of Gaussian elimination produced a lot of fractions along the way. Now less than 5 fourths of the population have problems with fractions, so if you like working with fractions then classic Gaussian elimination is perfectly feasible. And in fact we can't avoid fractions forever because sooner or later we'll run into a problem whose solutions are fractions or decimals. However there are some ways that we can delay using fractions for as long as possible, so we'll take a look at that next.