SOLUTION: I am reading my solutions and I don't understand how I get from one tableau to another. We pivoted around the second row and second column.
1 2 4 1 0 0 | 8
3 8 1 0 1
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-> SOLUTION: I am reading my solutions and I don't understand how I get from one tableau to another. We pivoted around the second row and second column.
1 2 4 1 0 0 | 8
3 8 1 0 1
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Question 475047: I am reading my solutions and I don't understand how I get from one tableau to another. We pivoted around the second row and second column.
1 2 4 1 0 0 | 8
3 8 1 0 1 0 | 10
-4 -32 1 0 0 1 | 0
to
1 0 15 4 -1 0 | 22
3 8 1 0 1 0 | 10
8 0 5 0 4 1 | 40 Answer by solver91311(24713) (Show Source):
Any row in a matrix, can be replaced by the sum of a constant times any other row plus row i. That is:
Actually, in more general terms, you can multiply both rows by constants if it makes sense to do so, thus:
1. Multiply Row 2 by 4: 12 32 4 0 4 0 | 40
2. Add the result of Step 1 to Row 3:
12 + (-4) = 8
32 + (-32) = 0
4 + 1 = 5
0 + 0 = 0
4 + 0 = 4
0 + 1 = 1
40 + 0 = 40
Hence: 8 0 5 0 0 4 1 | 40
3. Replace Row 3 with the results of Step 2.
Note: Just because you multiplied Row 2 by 4 in order to drive the 3rd Row 2nd Column element to zero, doesn't mean you have to replace Row 2 with the results of the multiplication.
John
My calculator said it, I believe it, that settles it
