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Having filled up with zeros the elements of column 1 below the pivot, now we
proceed to check the pivot at position (2,2). We find that the number 3 in
position (2,3) will be a better pivot, thus, we exchange columns 2 and 3 by
using: 2#3 ‚N@@@OK@@
Checking the pivot at position (2,2), we now find that the value of 25/8, at
position (3,2), is larger than 3. Thus, we exchange rows 2 and 3 by using:
2#3 L@RSWP
Now, we are ready to divide row 2 by the pivot 25/8, by using
³ 8/25™#2 L @RCI
Next, we eliminate the 3 from position (3,2) by using:
3\#2#3@RCIJ
Having filled with zeroes the position below the pivot, we proceed to check the
pivot at position (3,3). The current value of 2 is larger than ½ or 0, thus, we
keep it unchanged. We do divide the whole third row by 2 to convert the pivot
to 1, by using: 2Y3@RCI
Next, we proceed to eliminate the ½ in position (1,3) by using:
1 -1/16 1/2 41/16 0 1 0
03 2 -1 100
0 25/8 0 -25/8 0 0 1
1 -1/16 1/2 41/16 0 1 0
0 25/8 0 -25/8 0 0 1
03 2 -1 100
1 -1/16 1/2 41/16
010
010-1
001
032-1
100
1 -1/16 1/2 41/16
01 0
01 0 -1
00 1
00 2 2
10 0
1 -1/16 1/2 41/16
010
01 0 -1
001
00 1 1
100