by: prof. y.p. chiu1 extra -2 review of linear systems extra -2 review of linear systems prof. y....

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By: Prof. Y.P. Chiu 1 Extra -2 Extra -2 Review of Linear Systems Review of Linear Systems By: Prof. Y. Prof. Y. Peter Chiu Peter Chiu 9 / 9 / 2011 2011

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Page 1: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 1

Extra -2Extra -2

Review of Linear SystemsReview of Linear Systems

By: Prof. Y. Peter ChiuProf. Y. Peter Chiu

9 / 2011 9 / 2011

Page 2: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 2

§ . L 23 : Cramer’s Rule

A X ‧ = B X = B =

nnnn2n21n1

2n2n222121

1n1n212111

bxaxaxa

bxaxaxa

bxaxaxa

nnn2n1

2n2221

1n1211

aaa

aaa

aaa

A

1

2

n

X

X

X

n

2

1

b

b

b

Page 3: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 3

X = A-1B

xi =

11 21 n1

12 22 n2

1

2

1i 2i ni

n

1n 21 nn

1

2

n

A A A

A A A

A A A

A A A

b

b

A A A

A A A b

A A A

A A A

x

x

x

A

Ai

§ . L 23 : Cramer’s Rule

Page 4: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 4

Example 23-1

-2X1 + 3X2 - X3= 1

X1 + 2X2 - X3 = 4 X1=

-2X1 - X2 + X3 = -3

B →

X1 =

2

2

1

12

11

21

3

A

A

A

A

A 1i

22

4

2

1

1

1

1

2

3

3

4

1

2X

1

1 4

13

3

2 1

1

2 6

2 2

3

3 1

1 2 4

4X3

2

2 -1 8

2 2

B →B

Page 5: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 5

§ . L 24 :

If ≠ 0

Then A① -1 exist

② Linear System has nontrivial

solution. ( 非 0 解 )

③ rank A = n

④ The rows (columns) of A are

linearly independent.

A

Page 6: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 6

§ . L 26 : Gauss- Jordan reduction

'

''

''

333

223

113

ba00

ba10

ba01

"

''

''

3

223

113

b100

ba10

ba01

1

2

3

1 0 0 b ''

0 1 0 b "

0 0 1 b ''

Page 7: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 7

§ . L 25 : Gaussian Elimination 高斯消去法

上三角 (upper triangular)

3333231

2232221

1131211

baaa

baaa

baaa

3

223

11312

b'100

b'a10

b'aa1

'

''

3

2

113

b'100

b"010

b"0'a1

3

2

1

b'100

b"010

'b"001

Page 8: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 8

§ . L 27 : Homework # 1 X1+ X2 + 2X3 = -1

X1- 2X2 + X3 = -5

3X1+ X2 + X3 = 3

(a) Using Gaussian Elimination method to find solution.

(b) Using Gauss-Jordan reduction method.

(c) Using Cramer’s rule

# 2 2X1+ 4X2 + 6X3 = 2

X1 + 2X3 = 0

2X1+ 3X2 - X3 = -5

Using Cramer’s rule to solve it.

Page 9: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 9

# 3 Solve 3 X1- X2 = 3

# 4 Solve 2X1+ X2 +3X3 = 2

X1 + X3 = 1

# 5 Solve X1+ 2X2 +3X3 = 6

4X1 + X3 = 4

2X1+ 4X2 +6X3 = 11

§ . L 27 : Homework

Page 10: By: Prof. Y.P. Chiu1 Extra -2 Review of Linear Systems Extra -2 Review of Linear Systems Prof. Y. Peter Chiu By: Prof. Y. Peter Chiu 9 / 2011 9 / 2011

By: Prof. Y.P. Chiu 10

The EndThe End