In the Gauss-Jordan elimination, you start with a system of three equations and three unknowns:Then you convert it to this matrix: Then you use row operations and end up with a matrix like this: That is, you get 0's in the lower left three elements. Then you convert back to a system of equations: Then you do what is called "back-substitution": 1. Solve the bottom equation for z. 2. Substitute that value of z in the middle equation, and solve for y. 3. Substitute the values of y and z in the top equation and solve for x. ---------- To get a 0 where the 3 is on the middle row: Multiply the top row by -3 and the middle row by 7, and add them together: -------------- Replace the second row by that, leaving the rest as is To get a 0 where the 5 is on the bottom row: Multiply the top row by -5 and the bottom row by 7, and add them together: -------------- Notice that as it turns out, we can divide that through by -2, so we might as well do that too, and get: We replace the bottom row by that, leaving the rest as is To get a 0 where the 2 is on the bottom row: Take the middle row as it is. Multiply the bottom row by 25, and add them together: -------------- Notice that as it turns out, we can divide that through by 448, so we might as well do that too, and get: We replace the bottom row by that, leaving the rest as is Then we convert that back to this system of equations: or rather, Now we do what is called "back-substitution": The bottom equation is already solved for z. Substitute in the middle equation: Substitute and in the top equation: So the solution is Edwin