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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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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

While this problem has more variables than equations, it will be started like any other problem.


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.


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)  -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


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


Column 1 Complete

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


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.


Column 2

-150 in column 2 is the pivot element. We want to make the rest of the column 0’s


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


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


Column 2 Complete

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


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.


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


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


Column 3 Complete

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


Done with + - rows

You have the diagonal elements with values and 0’s in the rest of those columns


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)


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


Solution

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


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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