Gauss – Jordan Elimination Method: Example 2

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

elimination method

Slide 1

The system of linear equations

– 3x + 2y = 6

2x + 4y = 3

• What is the next step?

Slide 2

Convert to a matrix of coefficients

– 3x + 2y = 6

2x + 4y = 3– 3 2 6

2 4 3

Now circle the pivot number.

Slide 3

Pivot Number and Pivot Row

– 3 2 6

2 4 3

1. Recall that the row with the pivot number (circled number) is called the pivot row.

2. What is the next step?

Slide 4

1. 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.

2. 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.

3. Thus the matrix becomes:

Slide 5

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

1 – 2/3 – 2

2 4 3

1. The notation (– 1/3) R1 means to multiply all the values in row 1, as signified by the R1 , by the value (– 1/3) , which is the reciprocal of – 3.

2. Now what is the next step?

(– 1/3) R1

Slide 6

1. Change any values above and or below the pivot

value to a 0.

2. 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.

3. In this case we want to change the 2 (in the second row, first column) to a 0, so we take the second row and add it to ( – 2) times the values in the pivot row.

4. Notation: R2 + (– 2) R1 Slide 7

On a scratch piece of paper, do the following row operation: R2 + (– 2) R1

R2

(– 2 ) R1

2 4 3

– 2 4/3 4

0 16/3 7

1. (– 2) R1 means multiply (– 2) to the values in row 1. So the row

– 2 – 4/3 4 is a result of multiplying (– 2) to 1 – 2/3 – 2

2. The row of values 0 16/3 7 comes from adding the corresponding values in the two rows above, hence the addition symbol in the notation [2] + (– 2) [ 1 ].

3. Now since R2 is at the beginning of the statement R2 + (– 2) R1 , replace row 2 with the 0 16/3 7 values

4. Thus the new matrix will be the following:Slide 8

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

1 – 2/3 – 2

0 16/3 7

1. Now what is the next step?

[2] + (– 2)[ 1]

Slide 9

Change the pivot number

1 – 2/3 – 2

0 16/3 7

1. Since all the values below the pivot value of 1 are now zeros, the pivot value moves down the diagonal .

2. The pivot value is now 16/3 and the pivot row is the 0 16/3 7 row (i.e. row 2, or R2 ).

3. What is the next step?Slide 10

1. 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.

2. 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.

3. Thus the matrix becomes:

Slide 11

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

1 – 2/3 – 2

0 1 21/16

1. ( 3/16) R2 means that you multiply 3/16 to the values in row 2 (i.e. multiply 3/16 to 0, 16/3, and 7.

2. The 21/16 is from multiplying (3/16) to 7.

3. Now what is the next step?

( 3/16) R2

Slide 12

1. Change any values above and or below the pivot

value to a 0.

2. 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.

3. In this case we want to change the – 2/3 (in the first row, second column) to a 0, so we take the first row and add it to (2/3) times the values in the pivot row.

4. Notation: R1 + (2/3) R2 Slide 13

On a scratch piece of paper, do the following row operation: R1 + (2/3) R2

R1

(2/3) R2

1 – 2/3 – 2

0 2/3 7/8

1 0 – 9/8

1. (2/3) R2 means multiply (2/3) to the values in row 2. So the row

0 2/3 7/8 is a result of multiplying (2/3) to 0 , 1 and 21/16

2. The row of values 1 0 – 9/8 comes from adding the corresponding values in the two rows above.

3. Now since R1 is at the beginning of the statement R1 + (2/3) R2 , replace row 1 with the 1 0 – 9/8 values

4. Thus the new matrix will be the following:Slide 14

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

1 0 – 9/8

0 1 21/16

1. Now what is the next step?

R1 + (2/3) R2

Slide 15

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 16

Convert back to a system of equations

1x + 0y = – 9/8

0x + 1y = 21/16 1 0 – 9/8

0 1 21/16

Now simplify the system of equations.

Slide 17

1x + 0y = – 9/8

0x + 1y = 21/16

Thus the solution is ( – 9/8 , 21/16 )

Slide 18

Thus

x = – 9/8

y = 21/16