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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

Gauss-Jordan Method.

How To complete

Problem 2.2 # 57

Produced by E. Gretchen Gascon


1 set up the matrix in a spread sheet

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

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

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

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 Complete

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


5 keep row 2 fixed

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

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

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

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 Complete

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


Work on column 3

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

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

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 Complete

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


Done with rows

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

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

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

Solution

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


Evaluate the findings

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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