Gauss-Jordan Method.

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Gauss-Jordan Method. . How To complete Problem 2.2 # 29 Produced by E. Gretchen Gascon. The problem. Plan to solve. Step 1 – write a matrix with the coefficients of the terms and as the last column the constant equivalents.

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### Gauss-Jordan Method.

How To complete

Problem 2.2 # 29

Produced by E. Gretchen Gascon

Plan to solve
• Step 1 – write a matrix with the coefficients of the terms and as the last column the constant equivalents.
• Step 2 – use the Gauss-Jordan method to manipulate the matrix so that the solution will become evident
2. Keep Row1 fixed

Focus on the first row-first column element 4. We want to make the rest of the column #1 into 0’s.

To accomplish that we will perform the various row operation. Start with row 2.

What would be a LCM of 4 and 2? (This helps to define the formula to be used.

Ans: 4. Therefore you multiply 1 times 4 = 4, and 2 times 2 = 4. This is the formula 2*Elements of R2 Plus -1 times the Elements of R1

3. Replace Each Element of Row 2

6

2+4

-18-24

-42

4-4

0

-6

-2-4

Add 2*R2C1 -1*R1C1  R2C1, this would be 2(2) -1(4)  0

Replace all the cells in Row 2 with this formula, just change the column you are working in.

2*R2C2 -1*R1C2  R2C2, 2*R2C3 -1*R1C3  R2C3,

2(-1) - 1(4)  -2+ -4 -6 2(1) -1(-4)  2+46

2*R2C4 -1*R1C4  R2C4,

2(-9) -1(24)  -18-24 -42

4. Replace Each Element of Row 3

8+4

12

-16

-4+24

-4+4

-12-4

20

0

Add -4*R3C1 +*R1C1  R3C1, this would be -4(1) +(4)  0

Replace all the cells in Row 2 with this formula, just change the column you are working in.

-4*R3C2 +*R1C1  R3C2, -4*R3C3 +*R1C3  R3C3,

-4*R3C4 +*R1C4  R3C4,

Column 1 Complete

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

Column 2

-6 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

12

12-12

0

-12+12

0

12+0

72-84

-12

Add 3*R1C1 +2*R2C1  R1C1, this would be 3(4) +0 12

Replace all the cells in Row 1 with this formula, just change the column you are working in.

7. Replace Each Element of Row 3

0+0

0

0

12-12

-4

-16+12

-64

20-84

Add R3C1 +2*R2C1  R3C1, this would be 0 + 3(0)  0

Replace all the cells in Row 1 with this formula, just change the column you are working in.

Column 2 Complete

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

Work on column 3

-4 in column 3 remains, we want all the other elements in column 3 0’s.

We will accomplish this by using row operations, but the first row already has a 0 in the third column so we can skip that row operation.

9 Replace Each Element of Row 2

0+0

0

-12

-12+12

0

-276

-12+0

-84-192

Add 2*R2C1 +3*R3C1  R1C1, this would be (0) +3(0) 0

Replace all the cells in Row 1 with this formula, just change the column you are working in.

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

Two options:Option # 1 – covert the matrix back to an algebraic equation and solve

12x = -12 (from row #1)

x = -1

-12y = -276 (from row # 2)

y = 23

-4z = -64 (from row # 3)

z = 16

We can write these equations because if you remember col#1 was the coefficients of the x, col#2 the coefficients of the y and col#3 the coefficients of the z.

Option 2: Use the rules for Gauss-Jordan - Divide through by the left most element in each row.

12/12

1

-1

-12/12

1

-6/-6

-276/(-6)

23

1

-64/(-4)

16

-4/-4

R1/r1c1 r1/12

R2/r2c2  r2/(-6)

R3/r3c3 r3/(-4)

You are finished