Gaussian Elimination with Back Substitution

We will use Gaussian Elimination to solve the linear system

x1 2x1 3x1 + − 2x2 x2 + + − 3x3 x3 x3 = = = 9 8 3  .

The augmented matrix is

   1 2 3 2 −1 0 3 1 −1         9 8 3     

The Gaussian Elimination algorithm proceeds as follows:

1 2 3 2 −1 0 3 1 −1         9 8 3		−	1 0 0 2 −5 −6 3 −5 −10         9 −10 −24	(Row 1) (Row 2−2Row 1) (Row 3−3Row 1)
		−	1 0 0 2 1 −6 3 1 −10         9 2 −24	(Row 1) (−15Row 2) (Row 3)
		−	1 0 0 2 1 0 3 1 −4         9 2 −12	(Row 1) (Row 2) (Row 3+6Row 2)
		−	1 0 0 2 1 0 3 1 1         9 2 3	(Row 1) (Row 2) (−14Row 3)

We have brought the matrix to row-echelon form. The corresponding system

x1 + 2x2 x2 + + 3x3 x3 x3 = = = 9 2 3  

is easily solved from the bottom up:

x3=3 x2+3=2−x2=−1 x1+2(−1)+3(3)=9−x1=2  

Thus, the solution of the original system is x1=2x2=−1x3=3