Gaussian Elimination Of A 4x5 Matrix Steven T

Gaussian Elimination of a 4x5 Matrix A

A =

20 5 12 8 9
6 15 9 7 52
10 7 23 42 8
12 5 4 3 21

First we divide the first row by 20 to get a pivot of 1 at the A (1,1) spot:

A =

1 .25 .6 .4 .45
6 15 9 7 52
10 7 23 42 8
12 5 4 3 21

Then we need to get a 0 just below the first pivot. To do this we must multiply the first row by -6 and add it to the second row.

A =

1 .25 .6 .4 .45
0 13.5 5.4 4.6 49.3
10 7 23 42 8
12 5 4 3 21

Then we must get a 0 in the 3rd row of the first column underneath the first pivot. To do this we must multiply the first row by -10 and add it to the third row.

A =

1 .25 .6 .4 .45
0 13.5 5.4 4.6 49.3
0 4.5 17 38 3.5
12 5 4 3 21

Then we must get a 0 in the 4th row of the first column. To do this we must multiply the first row by -12 and add it to the 4th row.

A =

1 .25 .6 .4 .45
0 13.5 5.4 4.6 49.3
0 4.5 17 38 3.5
0 2 -3.2 -1.8 15.6

Now we have to get a pivot in the second column at the A (2,2) spot to do this we must divide the second row by 13.5.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 4.5 17 38 3.5
0 2 -3.2 -1.8 15.6

Then we have to get a 0 underneath the second pivot to do this we must multiply the 2nd row by -4.5 and add it to the 3rd row.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 0 15.2 547/15 -194/15
0 2 -3.2 -1.8 15.6

Then we must get a 0 underneath the second column in the forth row. To do this we must multiply the second row by -2 and add it to the 4th row.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 0 15.2 547/15 -194/15
0 0 -4 -67/27 218/27

Now we must get a pivot point of 1 at the A(3,3) spot of the matrix. We must divide the 3rd row by 15.2.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 -4 -67/27 218/27

Then we have to get a 0 under the 3rd pivot point. To do this we must multiply the 3rd row by 4 and add it to the 4th row.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 0 3650/513 2396/513

Now we must get a pivot point of 1 in the 4th column of the 4th row. To do this we must multiply the fourth row by the reciprical of 3650/513 which is 513/3650.

A =

1 .25 .6 .4 .45
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

To make this into a Row Echelon form matrix we must get 0's above the pivot as well. To do this we must divide the 2nd row by -4 and add it to the first row.

A =

1 0 .5 17/54 -25/54
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

Now we must get a 0 in the 3rd column of the first row to do this we must divide the 3rd row by -2 and add it to the first row.

A =

1 0 0 -3370/3809 -77/2052
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

Now we must get a 0 in the 4th column of the first row. To do this we must multiply the 4th row by 3370/3809 and add it to the first row.

A =

1 0 0 0 427/786
0 1 .4 46/135 493/135
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

Now we must get a 0 in the third colum of the second row. To do this we must multiply the third row by -2/5 and add it to the second row.

A =
1 0 0 0 427/786
0 1 0 -635/1026 2048/513
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

Now we must get a zero in the third column of the second row. To do this we must multiply the 4th row by 635/1026 and add it to the second row .

A =

1 0 0 0 427/786
0 1 0 0 14449/3285
0 0 1 547/228 -97/114
0 0 0 1 1198/1825

Now We must get a zero in the 4th column of the third row for this matrix to be in row echelon form to do this we must multiply the 4th row by -547/228 and add that to the 3rd row.

A =

1 0 0 0 427/786
0 1 0 0 14449/3285
0 0 1 0 -4427/1825
0 0 0 1 1198/1825

This matrix is now in Row Echelon Form!!!!! Finally!! lol
Hmm… In octave I type rref (A) and get my answers a bit off but I follow my steps and dont see anything wrong?
-Steven Trevino

page revision: 1, last edited: 17 Sep 2011 19:14