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

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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.

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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.

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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

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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

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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.

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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.

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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

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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

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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

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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.

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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.

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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

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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

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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.

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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

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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

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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

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

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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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