Download Gauss-Jordan Example 2

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
2 4
6
3
Now circle the pivot number.
Slide 3
Pivot Number and Pivot Row
–3 2
2 4
6
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:
(– 1/3) R1
1 – 2/3
2
4
–2
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?
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
1.
2.
3.
4.
2
–2
0
4
4/3
16/3
3
4
7
(– 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
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 ].
Now since R2 is at the beginning of the statement R2 + (– 2) R1 , replace row 2
with the 0 16/3 7 values
Thus the new matrix will be the following:
Slide 8
From the matrix in slide 6, the new matrix becomes:
[2] + (– 2)[ 1]
1.
1 – 2/3 – 2
0 16/3
7
Now what is the next step?
Slide 9
Change the pivot number
1 – 2/3
0 16/3
1.
2.
3.
–2
7
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 16/3 and the pivot row is the 0 16/3 7 row (i.e. row 2,
or R2 ).
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:
( 3/16) R2
1.
2.
3.
1 – 2/3 – 2
0
1 21/16
( 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.
The 21/16 is from multiplying (3/16) to 7.
Now what is the next step?
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.
4.
1
0
1
– 2/3
2/3
0
–2
7/8
– 9/8
(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
The row of values 1 0 – 9/8 comes from adding the corresponding
values in the two rows above.
Now since R1 is at the beginning of the statement R1 + (2/3) R2 , replace row 1
with the 1 0 – 9/8 values
Thus the new matrix will be the following:
Slide 14
From the matrix in slide 12, the new matrix becomes:
R1 + (2/3) R2
1.
1
0
0
1
– 9/8
21/16
Now what is the next step?
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
1
0
0 – 9/8
1 21/16
1x + 0y = – 9/8
0x + 1y = 21/16
Now simplify the system of equations.
Slide 17
Thus
1x + 0y = – 9/8
0x + 1y = 21/16
x = – 9/8
y = 21/16
Thus the solution is ( – 9/8 , 21/16 )
Slide 18