3x+2y+5z=1800 4x+y+3z+1450 2x+4y+z=1900 The method is easier if you don't try to get 1's on the diagonal, but just get three 0's in the lower left corner. That way you can keep everything whole numbers:4R1-3R2 -> R2 12 8 20 | 7200 -12 -3 -9 | -4350 -------------------- 0 5 11 | 2850 -2R1+3R3 -> R3 -6 -4 -10 | -3600 6 12 3 | 5700 -------------------- 0 8 -7 | 2100 8R2-5R3 -> R3 0 40 88 | 22800 0 -40 35 | -10500 -------------------- 0 0 123 | 12300 [Note that we can now easily get 1's on the diagonal from here, just by dividing each row by the first non-zero element in the row, like this: But why bother? Why not just leave all the elements as whole numbers? Why books and teachers insist that the first non-zero element in every row must be 1 is just plain dumb!!! Yes, I'm saying that teachers and books are all dumb to require that you make the first non-zero element in each row a 1 in Gaussian elimination. You certainly don't have to as you go!!!] I would just not bother making the first non-zero elements in each row be 1, and just take it as it is: Now the system is simply: 3x + 2y + 5z = 1800 5y + 11z = 2850 123z = 12300 Solving the bottom equation for z 123z = 12300 z = 100 Substituting z = 100 in the 2nd equation 5y + 11(100) = 2850 5y + 1100 = 2850 5y = 1750 y = 350 Substituting y = 350 and z = 100 in the first equation 3x + 2(350) + 5(100) = 1800 3x + 700 + 500 = 1800 3x + 1200 = 1800 3x = 600 x = 200 So the solution is (x,y,z) = (200,350,100) Since your teacher is probably one of the dumb-bunnies who requires that the first non-zero element in each row be 1, then just use the matrix above where we divided every row by its first non-zero element and got: and go from there, dealing with the fractions. This is just to satisfy the dumb mathematicians who think it's necessary to have 1's on the diagonal. It isn't at all! Edwin