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 2550751001200 30 60600 3020 30400 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 2550751001200 30 60600 3020 30400 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)  -300 5*R2C4 -6*R1C4  R2C4, 5*R2C5 -6*R1C5  R2C5 5(60) -6(100)  -300 5(600) -6(1200)  -4200 150-1500 150-300-150150-450-300300-600-300 3000-7200 -4200

5. 4. Replace Each Element of Row 3 2550751001200 0-150-300 -4200 3020 30400 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 150-1500100-300 100-450 150-650 2000-7200 -200-350-450 -5200

6. Column 1 Complete 2550751001200 0-150-300 -4200 0-200-350-450-5200 Column 1 – row 1 has a value and the rest of the column values are 0.

7. 5. Keep Row 2 fixed 2550751001200 0-150-300 -4200 0-200-350-450-5200 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 2550751001200 0-150-300 -4200 0-200-350-450-5200 -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 2550751001200 0-150-300 -4200 0-200-350-450-5200 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 75+0150-150225-300 300-300 3600-4200 750 -75 0-600

10. 7. Replace Each Element of Row 3 750-750-600 0-150-300 -4200 0-200-350-450-5200 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 0-0 -600+600 - 1050+1200 -1350+1200 -15600+16800 00 150-1501200

11. Column 2 Complete 750-750-600 0-150-300 -4200 00150-1501200 Column 2 – row 2 has a value and the rest of the column values are 0.

12. Work on column 3 750-750-600 0-150-300 -4200 00150-1501200 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 1 750-750-600 0-150-300 -4200 00150-1501200 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 150+0150 0+00 -150+150 00-150-150 -1200+1200 0

14. 9 Replace Each Element of Row 2 15000-1500 0 -300 -4200 00150-1501200 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 0-00 -150+0 -150 -300+300 0 -300-300 -600 -4200+2400 -1800

15. Column 3 Complete 15000-1500 0 0-600-1800 00150-1501200 Column 3 – row 3 has a value and the rest of the column values are 0.

16. Done with + - rows 15000-1500 0 0-600-1800 00150-1501200 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. 15000-1500 0 0-600-1800 00150-1501200 R1/r1c1  r1/150 R2/r2c2  r2/(-150) R3/r3c3  r3/(150) 150/150 1 -150/150 -150/-150 1 -600/-150 4 -1800/-150 12 150/150 1 -150/150 1200/150 8

18. You are finished 1000 010412 0018 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.