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Gauss-Jordan Method. How To complete Problem 2.2 # 57 Produced by E. Gretchen Gascon.

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Presentation on theme: "Gauss-Jordan Method. How To complete Problem 2.2 # 57 Produced by E. Gretchen Gascon."— Presentation transcript:

1 Gauss-Jordan Method. How To complete Problem 2.2 # 57 Produced by E. Gretchen Gascon

2 1. Set up the Matrix in a Spread sheet. Problem 2.2 #57 : R1 = Fiber, R2 = Protein, R3 = Fat C1 = Brand A, C2 = Brand B, C3 = Brand C, C4= Brand D, C5 = Totals While this problem has more variables than equations, it will be started like any other problem.

3 2. Keep Row1 fixed Focus on the first row-first column element 25. We want to make the rest of the column #1 into 0’s. To accomplish that we will perform the various row operation. What would be a LCM of 25 and 30? Ans: 150. Therefore you multiply 25 times 6 = 150, and 30 times 5 = 150.

4 3. Replace Each Element of Row Add 5*R2C1 -6*R1C1  R2C1, this would be 5(30) -6(25)  0 Replace all the cells in Row 2 with this formula, just change the column you are working in. 5*R2C2 -6*R1C1  R2C2, 5*R2C3 -6*R1C3  R2C3, 5(30) - 6(50)  -1505(30) -6(75)  *R2C4 -6*R1C4  R2C4, 5*R2C5 -6*R1C5  R2C5 5(60) -6(100)  (600) -6(1200) 

5 4. Replace Each Element of Row Add 5*R3C1 -6*R1C1  R3C1, this would be 5(30) -6(25)  0 Replace all the cells in Row 2 with this formula, just change the column you are working in. 5*R3C2 -6*R1C1  R3C2, 5*R3C3 -6*R1C3  R3C3, 5*R3C4 -6*R1C4  R3C4, 5*R3C5 -6*R1C5  R3C

6 Column 1 Complete Column 1 – row 1 has a value and the rest of the column values are 0.

7 5. Keep Row 2 fixed Focus on the second row-second column element We want to make the rest of the second column into 0’s. To accomplish that we will perform the various row operation.

8 Column in column 2 is the pivot element. We want to make the rest of the column 0’s

9 6. Replace Each Element of Row Add 3*R1C1 +R2C1  R1C1, this would be 3(25) +0  75 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 3*R1C2 +R2C2  R1C2, Add 3*R1C3 +R2C3  R1C3, Add 3*R1C4 +R2C4  R1C4, Add 3*R1C5 +R2C5  R1C

10 7. Replace Each Element of Row Add 3*R3C1 -4R2C1  R3C1, this would be 3(0) +0  0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 3*R3C2 -4R2C2  R3C2, Add 3*R3C3 -4R2C3  R3C3, Add 3*R3C4 +-4R2C4  R3C4, Add 3*R3C5 -4R2C5  R3C

11 Column 2 Complete Column 2 – row 2 has a value and the rest of the column values are 0.

12 Work on column in column 3 remains, we want all the other elements in column 3 0’s. We will accomplish this by using row operations.

13 8 Replace Each Element of Row Add 2*R1C1 +R3C1  R1C1, this would be 2(75) +0  150 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 2*R1C2 +R3C2  R1C2, Add 2*R1C3 +R3C3  R1C3, Add 2*R1C4 +R3C4  R1C4, Add 2*R1C5 +R3C5  R1C

14 9 Replace Each Element of Row Add R2C1 +2*R3C1  R1C1, this would be (0) +2(0)  0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add R2C2 +2*R3C2  R1C2, Add R2C3 +2*R3C3  R1C3, Add R2C4 +2*R3C4  R1C4, Add R2C5 +2*R3C5  R1C

15 Column 3 Complete Column 3 – row 3 has a value and the rest of the column values are 0.

16 Done with + - rows You have the diagonal elements with values and 0’s in the rest of those columns

17 Divide through by the left most element in each row R1/r1c1  r1/150 R2/r2c2  r2/(-150) R3/r3c3  r3/(150) 150/ / / / / / / /150 8

18 You are finished At this point you can read the answer from the matrix. A – D = 0 B + 4D = 12 C – D = 8 ABCD Unfortunately with this problem there is not a unique solution

19 Solution Because this problem does not have a unique solution we solve all variables in terms of D

20 Evaluate the Findings All variables must be >= 0 Substitute various values for D If D = 0, then A=0, B=12, C=8, D =0 If D = 1, then A=1, B=8, C=9, D =1 If D = 2, then A=2, B=4, C=10, D =2 If D = 3, then A=3, B=0, C=11, D =3 If D = 4, then A=4, B=-4, C=12, D =4 There is no use to go further, because All the variables must be >= 0 and at D=4, B is negative.


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