ch8 Numerical Technique

c 1 0 0 5e 5 f = 0 5 0 ⋅ A = E2 A h i 0 0 1 b
Ch8_6
1 2 3 2 − 1 0 0 1 3 1 0 0 0 1 2
≈ R1 + (−2) R 2 R 2 + (−3) R3
Comparison of Gauss-Jordan and Gaussian Elimination
Count of Operations for n × n system with Unique Solution Number of Multiplications Gauss-Jordan Gaussian elimination n3 n 2 n3 + ≈ ( for large n ) 2 2 2 n3 n n3 2 +n − ≈ 3 3 3 Number of Additions n3 n n3 − ≈ 2 2 2 n 3 n 2 5n n 3 + − ≈ 3 2 6 3
upper triangular matrix
Ch8_8
Example 1 Solving the following system of equations, which has a triangular matrix of coefficients. 2x =8
1
x +3x2 1
= −2
Solution
Ch8_3
The corresponding system of equation is x1 + 2 x2 + 3 x3 + 2 x4 = −1 x3 + 3 x4 = 1 x4 = 2 We get x3 + 3(2) = 1 x3 = −5 Substituting x4 = 2 and x3 = −5 into the first equation, x1 + 2 x2 + 3(−5) + 2(2) = −1 x1 + 2 x2 = 10 x1 = −2 x2 + 10 Let x2 = r. The system has many solutions. The solutions are x1 = −2r + 10, x2 = r , x3 = −5, x4 = 2
4x +5x2 −x3 =3 1
By forward substitution: 1st equations gives 2 x1 = 8 ⇒ x1 = 4 2nd equation gives 3rd equation gives
x1 + 3x2 = −2 ⇒ 4 + 3x2 = −2 ⇒ x2 = −2
1 0 0 I 3 = 0 1 0 0 0 1
R2 ↔ R3
1 0 0 E1 = 0 0 1 0 1 0 1 0 0 E 2 = 0 5 0 0 0 1 1 0 0 E3 = 2 1 0 0 0 1
Ch8_7
8.2 The Method of LU Decomposition
0 2 0 3 − 1 0 5 2 7 4 0 − 2 0 0 0 8 8 0 0 0 − 3 1 0 −4 2 0 0 7 2 2 5 9
lower triangular matrix Ax=y ⇒ A=LU …
Ch8_10
Definition
Let A be a square matrix that can be factored into the form A = LU, where L is a lower triangular matrix and U is an upper triangular matrix, This factoring is called an LU decomposition of A. (Not every matrix has an LU decomposition, and when it exists, it is not unique.)
Ch8_13
5R2
R2+ 2R1
Elementary Matrices（複習第二章)
一個矩陣做 elementary row operation， 相當於在左邊乘一個對應的 elementary matrix。
a b R2 ↔ R3 g h d e
c 1 0 0 i = 0 0 1 ⋅ A = E1 A f 0 1 0
Ch8_4
Example 2
Solving the following system of linear equations using the method of Gaussian elimination, performing back substitution using matrices. x +2x2 +3x3 +2x4 = − 1 1 −x −2x2 −2x3 + x4 = 2 1 2x +4x2 +8x3 +12x4 = 4 1 Solution We arrive at the echelon form as in the previous example. 2 3 2 − 1 1 1 2 3 2 − 1 − 1 − 2 − 2 1 2 ≈ L ≈ 0 0 1 3 1 0 0 0 1 2 4 8 12 4 2
Ch8_11
Method of LU Decomposition
Let AX = B be a system of n equations in n variables, where A has LU decomposition A = LU. ⇒ LUX = B ⇒ two subsystems: UX = Y (upper triangular) and LY = B (lower triangular)
Linear Algebra
Chapter 8
Numerical Technique
大葉大學 資訊工程系 黃鈴玲
8.1 Gaussian Elimination
Definition
A matrix is in echelon form if 1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix. 2. The first nonzero element of each row is 1. This element is called a leading 1. . 3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row. (This implies that all the elements below a leading 1 are zero.) Reduced echelon form與echelon from的差異： echelon form 的 leading 1 上面的數字不必為零
Ch8_2
Example 1
Solving the following system of linear equations using the method of Gaussian elimination. x +2x2 +3x3 +2x4 = − 1 1 −x −2x2 −2x3 + x4 = 2 1 2x +4x2 +8x3 +12x4 = 4 1 Solution Starting with the augmented matrix, create zeros below the pivot in the first column. 2 3 2 − 1 ≈ 1 1 2 3 2 − 1 − 1 − 2 − 2 1 2 R 2 + R1 0 0 1 3 1 2 4 8 12 4 R3 + (−2) R1 0 0 2 8 6 At this stage, we create a zero only below the pivot. 1 2 3 2 − 1 ≈ 1 2 3 2 − 1 ≈ 0 0 1 3 1 1 0 0 1 3 1 R3 + (−2) R 2 2 R 3 0 0 0 1 2 0 0 0 2 4 Echelon form We have arrived at the echelon form.
2x −x2 +4x3 =0 1 x2 −x3 = 4 3x3 = −6
Solution By back substitution:
3rd equation gives 3 x3 = −6 ⇒ x3 = −2 2nd equation gives x2 − x3 = 4 ⇒ x2 + 2 = 4 ⇒ x2 = 2 1st equations gives 2 x1 − x2 + 4 x3 = 0 ⇒ 2 x1 − 2 − 8 = 0 ⇒ x1 = 5 The solution is x1 = 5, x2 = 2, x3 = −2.
Echelon form
This marks the end of the forward elimination of variables from equations. We now commence the back substitution using matrices.
Ch8_5
1 2 3 0 − 5 0 0 1 0 − 5 0 0 0 1 2 ≈ 1 2 0 0 10 R1 + (−3) R 3 0 0 1 0 − 5 0 0 0 1 2 This matrix is the reduced echelon form of the original augmented matrix. The corresponding system of equations is x1 + 2 x2 = 10 x3 = −5 x4 = 2 Let x2 = r. We get same solution as previously, x1 = −2r + 10, x2 = r , x3 = −5, x4 = 2