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Gauss-Jordan Method.

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

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Gauss-Jordan Method.

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  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 2 150-150 0 150-300 -150 150-450 -300 300-600 -300 -4200 3000-7200 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)  -150 5(30) -6(75)  -300 5*R2C4 -6*R1C4  R2C4, 5*R2C5 -6*R1C5  R2C5 5(60) -6(100)  -300 5(600) -6(1200)  -4200

  5. 4. Replace Each Element of Row 3 -200 100-450 -350 -450 150-150 0 100-300 150-650 -5200 2000-7200 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  R3C5

  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 -150. 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 2 -150 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 1 300-300 75+0 75 150-150 0 225-300 0 -600 -75 3600-4200 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  R1C5

  10. 7. Replace Each Element of Row 3 0-0 0 0 -600+600 -1050+1200 150 -1350+1200 -150 1200 -15600+16800 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  R3C5

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

  12. Work on column 3 150 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 1 150+0 150 0+0 0 -150+150 0 0-150 -150 0 -1200+1200 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  R1C5

  14. 9 Replace Each Element of Row 2 0-0 0 0 -1800 -150 -300+300 -600 -150+0 -300-300 -4200+2400 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  R1C5

  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. 1 150/150 -1 -150/150 1 4 12 -150/-150 -600/-150 -1800/-150 -1 8 1 -150/150 1200/150 150/150 R1/r1c1 r1/150 R2/r2c2  r2/(-150) R3/r3c3 r3/(150)

  18. You are finished A B C D At this point you can read the answer from the matrix. A – D = 0 B + 4D = 12 C – D = 8 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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