Download Section 2.2

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Section 2.2
Systems of Liner Equations:
Unique Solutions
The Gauss-Jordan Elimination Method
Operations
1. Interchange any two equations.
2. Replace an equation by a nonzero
constant multiple of itself.
3. Replace an equation by the sum of
that equation and a constant
multiple of any other equation.
Ex. Solve the system
2x  y  z  3
step
1
x yz  2
2y  2 z  2
3y  z  1
2
x y z  2
 x  3 y  3z  0
x yz 2
y  z 1
3 y  z  1
Row 1 (r1)
Row 2 (r2)
Row 3 (r3)
Replace r2 with [r1 + r2]
Replace r3 with [–2(r1) + r3]
Replace r2 with ½(r2)
x yz  2
3
y  z 1
2 z  4
Replace r3 with [–3(r2) + r3]
x yz 2
4
y  z 1
z  2
x y
5
y
Replace r3 with ½(r3)
4
Replace r1 with [(–1)r3 + r1]
 1
Replace r2 with [r2 + r3]
z  2
3
x
6
y
Replace r1 with [r2 + r1]
 1
z  2
So the solution is (3, –1, –2)
Augmented Matrix
*Notice that the variables in the preceding
example merely keep the coefficients in line.
This can also be accomplished using a matrix.
A matrix is a rectangular array of numbers
System
x y z  2
 x  3 y  3z  0
2x  y  z  3
Augmented matrix
 1 1 1 2 



1
3

3
0


 2 1 1 3 
coefficients
constants
Row Operation Notation:
1.Interchange row i and row j
Ri  R j
2.Replace row j with c times row j
cR j
3.Replace row i with the sum of row i and
c times row j
Ri  cR j
Ex. Last example revisited:
Matrix
System
x y z  2
 x  3 y  3z  0
 1 1 1 2 



1
3

3
0


 2 1 1 3 
2x  y  z  3
x yz  2
2y  2 z  2
3y  z  1
R2  R1
R3  ( 2) R1
1 1 1 2 


0
2

2
2


0 3 1 1
x yz 2
y  z 1
1
R
2 2
3 y  z  1
x yz  2
y  z 1
2 z  4
R3  ( 3) R2
x yz 2
y  z 1
z  2
1
R
2 3
 1 1 1 2 


0 1 1 1 
0 3 1 1
1 1 1 2 


0 1 1 1 
0 0 2 4 
1 1 1 2 


0
1

1
1


0 0 1 2 
x y
y
4
 1
R1  ( 1) R3
R2  R3
z  2
3
x
y
 1
R1  R2
z  2
This is in Row- Reduced
Form
1 1 0 4 


0 1 0 1
0 0 1 2 
1 0 0 3 


0
1
0

1


0 0 1 2 
Row–Reduced Form of a Matrix
1. Each row consisting entirely of zeros lies
below any other row with nonzero entries.
2. The first nonzero entry in each row is a 1.
3. In any two consecutive (nonzero) rows, the
leading 1 in the lower row is to the right of
the leading 1 in the upper row.
4. If a column contains a leading 1, then the
other entries in that column are zeros.
Row–Reduced Form of a Matrix
Row-Reduced Form
Non Row-Reduced Form
1 0 0 3 


0 1 0 1
0 0 1 2 
1 0 0 9 


0 0 1 4 
0 1 0 2 
1 0 0 8 


0 1 0 4 
0 0 0 0 
R2 , R3
switched
Must
be 0
1 0 5 1 


0 1 0 3 
0 0 1 5
Unit Column
A column in a coefficient matrix where one of the
entries is 1 and the other entries are 0.
1 0 5 1 


0
1
0
3


0 0 1 5
Unit columns
Not a Unit
column
Pivoting – Using a coefficient to transform a
column into a unit column
 1 1 1 2 



1
3

3
0


 2 1 1 3 
1 1 1 2 


0
2

2
2


0 3 1 1
This is called pivoting on the 1 and it is
circled to signify it is the pivot.
Gauss-Jordan Elimination Method
1. Write the augmented matrix
2. Interchange rows, if necessary, to obtain a
nonzero first entry. Pivot on this entry.
3. Interchange rows, if necessary, to obtain a
nonzero second entry in the second row. Pivot
on this entry.
4. Continue until in row-reduced form.
Related documents