Multivariable linear systems

 The following system is said to be in row-echelon form, which means that it has a “stair-step” pattern with leading coefficients of 1. Using Back-Substitution in Row- Echelon Form

Back-Substitution Equation 1 Equation 2 Equation 3

Back-Substitution Equation 1 Equation 2 Equation 3 From equation 3, you know the value of z. To solve for y, substitute z=2 into Equation 2 to obtain

Back-Substitution Finally, substitute y=-1 and z=2 into equation 1 to obtain. The solution is x=1, y=-1, and z=2, which can be written as the ordered triple (1, -1, 2). Check this in the original system of equations.

Practice

 Two systems of equations are equivalent if they have the same solution set. To solve a system that is not in row-echelon form, first convert it to an equivalent system that is in row-echelon form by using own or more of the elementary row operations shown in the next example. This process is called Gaussian elimination. Gaussian Elimination

 Interchange two equations.  Multiply own of the equations by a nonzero constant.  Add a multiple of one equation to another equation. Elementary Row Operations for Systems of Equations

Example Equation 1 Equation 2 Equation 3

Example Continued Adding the first equation to the second equation produces a new second equation.

Example Continued Adding -2 times the first equation to the third equation produces a new third equation.

 Now that all but the first x have been eliminated from the first column, go to work on the second column. (You need to eliminate y from the third equation. Example Continued Adding the second equation to the third equation produces a new third equation.

 Finally, you need a coefficient of 1 for z in the third equation. Example Continued

 You can conclude that the solution is x=1, y=-1, and z=2, written as (1, -1, 2). Example Continued

 Solve the system of linear equations. Try this…