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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. The system of linear equations. 4x – 3y = 7 3x – 2y = 6. What is the next step?. Convert to a matrix of coefficients. 4x – 3y = 7
Gauss – Jordan Elimination Method: Example 1

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Slide 1

Gauss – Jordan Elimination Method: Example 1

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

Slide 2

The system of linear equations

4x – 3y = 7

3x – 2y = 6

  • What is the next step?

Slide 3

Convert to a matrix of coefficients

4x – 3y = 7

3x – 2y = 6

4 – 3 7

3 – 2 6

Now circle the pivot number.

Slide 4

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?

Slide 5

  • 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:

Slide 6

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?

Slide 7

  • 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

Slide 8

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:

Slide 9

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

1 – 3/4 7/4

0 1/4 3/4

R2 + (– 3) R1

  • Now what is the next step?

Slide 10

Change the pivot number

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?

Slide 11

  • 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.

  • Thus the matrix becomes:

Slide 12

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

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?

Slide 13

  • 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

Slide 14

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:

Slide 15

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

1 0 4

0 1 3

R1 + (3/4) R2

  • Now what is the next step?

Slide 16

Convert the matrix back to a system of equations

  • 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.

Slide 17

Convert back to a system of equations

1x + 0y = 3

0x + 1y = 4

1 0 3

0 1 4

Now simplify the system of equations.

Slide 18

Thus

x = 3

y = 4

1x + 0y = 3

0x + 1y = 4

Thus the solution is ( 3 , 4 )


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