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Gauss – Jordan Elimination Method: Example 1PowerPoint Presentation

Gauss – Jordan Elimination Method: Example 1

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### Gauss – Jordan Elimination Method: Example 1

Solve the following system of linear equations using the Gauss-Jordan elimination method

Convert to a matrix of coefficients

4x – 3y = 7

3x – 2y = 6

4 – 3 7

3 – 2 6

Now circle the pivot number.

Pivot Number and Pivot Row

4 – 3 7

3 – 2 6

- Recall that the row with the pivot number (circled number) is called the pivot row.
- What is the next step?

- Change the pivot number to a 1 by multiplying the pivot number, and all the other numbers in the pivot row,by the reciprocal of the circled number.
- Remember that when you change a pivot number to a 1, you use the second elementary row operation; Multiply an equation by a nonzero value.
- Thus the matrix becomes:

From the matrix in slide 4, the new matrix becomes:

1 – 3/4 7/4

3 – 2 6

(1/4) R1

- The notation (1/4)R1 means to multiply all the values in row 1, as signified by the R1, by the value of (1/4) , which is the reciprocal of 4.
- Now what is the next step?

- Change any values above and or below the pivot value to a 0.
- Do this by multiplying the pivot row by the opposite number (i.e. change the sign of the number) that you want to change to a 0.
- In this case we want to change the 3 (in the second row, first column) to a 0, so we take the second row and add it to ( – 3) times the values in the pivot row.
- Notation: R2 + (– 3) R1

On a scratch piece of paper, do the following row operation: R2 + (– 3) R1

R2

(– 3 ) R1

3 – 2 6

– 3 9/4 –21/4

0 1/4 3/4

- 1. (– 3) R1 means multiply (– 3) to the values in row 1. So the row
- – 3 9/4 – 21/4 is the result of multiplying (– 3) to 1 , – 3/4 and 7/4.
- The row of values 0 1/4 3/4 comes from adding the corresponding values in the two rows above, hence the addition symbol in the notation R2 + (– 3) R1 .
- Now since R2 is at the beginning of the statement R2 + (– 3) R1 , replace row 2 with the 0 1/4 3/4 values
- Thus the new matrix will be the following:

From the matrix in slide 6, the new matrix becomes: R

1 – 3/4 7/4

0 1/4 3/4

R2 + (– 3) R1

- Now what is the next step?

Change the pivot number R

1 – 3/4 7/4

0 1/4 3/4

- Since all the values below the pivot value of 1 are now zeros, the pivot value moves down the diagonal .
- The pivot value is now 1/4 and the pivot row is the 0 1/4 3/4 row (i.e. row 2, or R2 ).
- What is the next step?

- Change the pivot number to a 1 by multiplying the pivot Rnumber, and all the other numbers in the pivot row,by the reciprocal of the circled number.
- Remember that when you change a pivot number to a 1, you use the second elementary row operation; Multiply an equation by a nonzero.
- Thus the matrix becomes:

From the matrix in slide 10, the new matrix becomes: R

1 – 3/4 7/4

0 1 3

( 4) R2

- ( 4) R2 means that you multiply 4 to the values in row 2 (i.e. multiply 4 to
- 0 1/4 and ¾.
- The 3 is from multiplying 4 to (3/4).
- Now what is the next step?

- Change any values above and or below the pivot value to a 0.
- Do this by multiplying the pivot row by the opposite number (i.e. change the sign of the number) that you want to change to a 0.
- In this case we want to change the – 3/4 (in the first row, second column) to a 0, so we take the first row and add it to (3/4) times the values in the pivot row.
- Notation: R1 + (3/4) R2

On a scratch piece of paper, do the following row operation: R1 + (3/4) R2

R1

(3/4) R2

1 – 3/4 7/4

0 3/4 9/4

1 0 4

- 1. (3/4) R2 means multiply (3/4) to the values in row 2. So the row
- 0 3/4 9/4 is a result of multiplying (3/4) to 0 , 1 and 3
- The row of values 1 0 4 comes from adding the corresponding values in the two rows above.
- Now since R1 is at the beginning of the statement R1 + (3/4) R2 , replace row 1 with the 1 0 4 values
- Thus the new matrix will be the following:

From the matrix in slide 12, the new matrix becomes: R

1 0 4

0 1 3

R1 + (3/4) R2

- Now what is the next step?

Convert the matrix back to a system of equations R

- Now that there are 1’s on the diagonals (from top left corner to the bottom right corner) and 0’s above and/or below the 1’s, then convert the matrix back to the system of linear equations.

Convert back to a system of equations R

1x + 0y = 3

0x + 1y = 4

1 0 3

0 1 4

Now simplify the system of equations.

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