sec 3.2 matrices and gaussian elemination coefficient matrix 3 x 3 coefficient matrix 3 x 3...
TRANSCRIPT
(eq3) 23972
(eq2) 20783
(eq1) 4 2
zyx
zyx
zyxExample
Sec 3.2 Matrices and Gaussian Elemination
972
783
121
23972
20783
4121
Coefficient Matrix3 x 3
Augmented Coefficient Matrix3 x 4
n) (eq aa aa
2) (eq aa aa
1) (eq aa aa
1nn3n32n21n1
12n323222121
11n313212111
bxxxx
bxxxx
bxxxx
n
n
n
System
Linear
Sec 3.2 Matrices and Gaussian Elemination
nnnn
n
n
aaa
aaa
aaa
21
22221
11211
nnnnn
n
n
baaa
baaa
baaa
21
222221
111211
Coefficient Matrixn x n
Augmented Coefficient Matrixn x (n+1)
Elementary Row OperationsMultiply one equation by a nonzero constant1
Equ(i) * C
Interchange two equations2
Equ(j) Equ(i)
Add a constant multiple of one equation to another equation
3location Equ(j) Equ(j) Equ(i) * C
Multiply one row by a nonzero constant1 iR * C
Interchange two rows
2ji R R
Add a constant multiple of one row to another row
3ji R R * C
(eq3) 11892
(eq2) 5 23
(eq1) 221083
zyx
zyx
zyxExample
1)Extra HW2)Problem Session3)Quiz 2 Stat4)Chapter 1 Summ
How to solve any linear system
Use sequence of elementary row operations Triangular
system
*
*
*
z
y
xUse back substitu
tion
****
****
****
*100
**10
***1
How to solve any linear system
****
****
****
****
****
***1
***0
***0
***1
***0
**10
***1
**00
**10
***1
*100
**10
***1
(eq3) 23972
(eq2) 20783
(eq1) 4 2
zyx
zyx
zyxExample
(-3) R1 + R2
(-2) R1 + R3
(-3) R2 + R3
23972
20783
4121Augmented Matrix
23972
20783
4121
15730
8420
4121(1/2) R2
3100
4210
4121
15730
4210
4121
Definition: (Row-Equivalent Matrices)
A and B are row equivalent if B can be obtained from A by a finite sequence of elementary row operations
A
BExample A and B are row equivalent
A is the augmented matrix of sys(1)B is the augmented matrix of sys(2)
The
orem
1:
A and B are row equivalent&
sys(1) and sys(2) have same solution
Echelon Matrix
000000
430120
000000
020000
110101
zero row
Example How many zero rows
000000
000000
100000
021020
110101
010000
000001
100000
021020
110101
Echelon Matrix
non-zero row
Example
1) How many non-zero rows
2) Find all leading entries
000000
000000
100000
021020
110101
010000
000001
100000
021020
110101
000000
430120
000000
020000
110101
leading entry The first (from left) nonzero element in each nonzero row
0000
1120
1101
Echelon Matrix
Def: A matrix A in row-echelon form if
1) All zero rows are at the bottom of the matrix
2) In consecutive nonzero rows the leading in the lower row appears to the right of the leading in the higher row
1 5 0 2
0 1 0 1
0 0 0 0
A
1
1 5 0 2
0 2 0 1
0 0 0 0
A
2
1 5 0 2
0 0 1 1
0 1 0 1
A
3
1 5 0 2
0 0 0 0
0 1 0 1
A
1100000
0120000
1010000
2020010
1 5 0 2
0 1 0 1
0 0 0 0
A
1 0 5 2
0 1 0 1
0 0 0 0
B
000000
001000
100000
020010
010001
Echelon Matrix
2326542
2121363
1513163
Transform each augmented matrix to echelon form. Then use back substitution to solve the system
Def: A matrix A in reduced-row-echelon form if
1) A is row-echelon form
2) All leading entries = 1
3) A column containing a leading entry 1 has 0’s everywhere else
1 5 0 2
0 1 0 1
0 0 0 0
A
1 0 5 2
0 1 0 1
0 0 0 0
B
000000
001000
100000
020010
010001
Reduced Echelon Matrix
1) A row-echelon form
2) Make All leading entries = 1 (by division)
3) Use each leading 1 to clear out any nonzero elements in its column
1 5 0 2
0 1 0 1
0 0 0 0
A
1 0 5 2
0 1 0 1
0 0 0 0
B
000000
001000
100000
020010
010001
Echelon Matrix Reduced Echelon Matrix
4200000
9603300
4311211
Solving Linear System
Gaussian Elimination Method:
1 * * *
* * * *
* * * *
Gauss-Jordan Elimination Method:
A row-echelon form back subsutition
A reduced-row-echelon form
Solve:
1 2 3
1 2 3
1 2 3
2 6 7
2 1
5 7 4 9
x x x
x x x
x x x
:Example
A1 * * *
0 * * *
0 * * *
1 * * *
0 1 * *
0 * * *
1 * * *
0 1 * *
0 0 * *
1 * * *
0 1 * *
0 0 1 *
Row-echelon form
Reduced Row-echelon form
1 * * *
0 1 * *
0 0 1 *
1 * 0 *
0 1 0 *
0 0 1 *
1 0 0 *
0 1 0 *
0 0 1 *