Gaussian Elimination with scaled partial pivoting

æ ç ç ç ç è    3 1 2 9 |   5    2 1 2 1 | -2    6 1 1 8 |   2 -3 0 2 1 |   2 ö ÷ ÷ ÷ ÷ ø    row scales    æ ç ç ç ç è 9 2 8 3 ö ÷ ÷ ÷ ÷ ø    pivot order æ ç ç ç ç è 1 2 3 4 ö ÷ ÷ ÷ ÷ ø

Determine the pivot row by taking the row corresponding to the largest of [3/9, 2/2, 6/8, 3/3]. Row 2 is the pivot row that will be used to eliminate the first variable from equations 1,3, and 4. The multipliers are: [-3/2,X,-6/2,3/2] and the result of the elimination is:

æ ç ç ç ç è 0 -1/2 -1 15/2 |    8 2 1    2 1 | -2 0 -2 -5 5 |    8 0 3/2    5 5/2 | -1 ö ÷ ÷ ÷ ÷ ø    pivot order æ ç ç ç ç è 2 1 3 4 ö ÷ ÷ ÷ ÷ ø

Determine the next pivot by taking the row corresponding to the largest of [[(1/2)/9], X,2/8,[( 3/2)/4]] = [1/18, X, 1/4, 3/8]. Row 4 is the pivot row that will be used to eliminate the second variable from equations 1 and 3. The multipliers are: [1/3,X,4/3,X] and the result of the elimination is:

æ ç ç ç ç è 0 0 2/3 25/3 | 23/3 2 1 2 1 | -2 0 0 5/3 25/3 | 20/3 0 3/2 5 5/2 | -1 ö ÷ ÷ ÷ ÷ ø    pivot order æ ç ç ç ç è 2 4 3 1 ö ÷ ÷ ÷ ÷ ø

Determine the next pivot as the row corresponding to the largest of [2/27, X, 5/24, X]. Row 3 is the pivot row that will be used to eliminate the third variable from equation 1. The multiplier is: [-2/5,X,X,X] and the result of the elimination is:

æ ç ç ç ç è 0 0 0 5 | 5 2 1 2 1 | -2 0 0 5/3 25/3 | 20/3 0 3/2 5 5/2 | -1 ö ÷ ÷ ÷ ÷ ø    pivot order æ ç ç ç ç è 2 4 3 1 ö ÷ ÷ ÷ ÷ ø

Back Substitution

æ ç ç ç ç è 0 0 0 5 | 5 2 1 2 1 | -2 0 0 5/3 25/3 | 20/3 0 3/2 5 5/2 | -1 ö ÷ ÷ ÷ ÷ ø    pivot order æ ç ç ç ç è 2 4 3 1 ö ÷ ÷ ÷ ÷ ø

The pivot order determines the order in which equations will be solved. The equations are taken in the reverse of the pivot order.

5z = 5 Þ z=1 (1) 5 3 y + 25 3 z = 20/3 Þ y = (20/3-25/3)/(5/3)=-1 (3) 3 2 x + 5 y + 5 2 z = -1 Þ x=( -1 +5 -5/2)/(3/2)=1 (4) 2 w + x + 2 y +z = -2 Þ w=( -2 -1 +2 -1 )/2=-1 (2)

The determinant

There were no row interchanges so the determinant of the above matrix is the same as that of the original matrix.

5 (-1)1+4 (5/3) (-1)2+3 (3/2) (-1)2+2 2 =25

We could also note that the pivot order is 2,4,3,1 and the number of interchanges required to restore the natural order is 4.

(5)(5/3)(3/2)(2) =25