Using the column principal element elimination method to solve the ax=b of the equations respectively, and realize with MATLAB program (most effective version)

Numerical analysis often involves using the MATLAB program to solve the equations by using the column principal element elimination method respectively Ax=b

The specific methods and codes are illustrated by the following equation (3x3 matrix):

Using the column principal element elimination method to solve the equation group ax=b, the MATLAB program is used to realize:

(1)

1, MATLAB code to realize the solution of the equation can be divided into Two kinds , one is Entry-level , just simply calculate the problem, the second is a General-Purpose code , can realize many of the 3x3 matrix equation solution, write well later only need to change the elements in different matrices to calculate the corresponding solution, need to be established on the basis of familiarity with MATLAB, as follows:

The first code implementation-entry level:

A=[3.01,6.03,1.99,;1.27,4.16,-1.23,;0.987,-4.81,9.34]

a1=[3.01,6.03,1.99,1;1.27,4.16,-1.23,1;0.987,-4.81,9.34,1]

B1=A1 (1,1:4)

C1=A1 (2,1:4)

D1=A1 (3,1:4)

E1=-1.27/3.01*b1+c1

F1=-0.987/3.01*b1+d1

P1=e1 (ON)

Q1=F1 (ON)

if (ABS (p1) >=abs (Q1))

A1=p1

A2=q1

Ff1=e1

Ee1=f1

Else

A1=q1

A2=p1

Ff1=f1

Ee1=e1

End

G1=-a2/a1*ff1+ee1

H1=[e11; FF1; G1]

J1=H1 (1:3,1:3)

B1=H1 (1:3,4)

X1=j1\b1

The second type of code is implemented as follows-proficiency in the general level:

A=[3.01,6.03,1.99,;1.27,4.16,-1.23,;0.987,-4.81,9.34]

a1=[3.01,6.03,1.99,1;1.27,4.16,-1.23,1;0.987,-4.81,9.34,1]

B1=A1 (1,1:4)

C1=A1 (2,1:4)

D1=A1 (3,1:4)

F1=A1 (a)

F2=A1 (2,1)

F3=A1 (3,1)

if (ABS (F1) >=abs (F2))

if (ABS (F1) >=abs (F3))

F11=f1

E11=b1

F22=f2

E12=c1

F33=f3

E13=d1

Else

F11=f3

E11=d1

F22=f1

E12=b1

F33=f2

E13=c1

End

End

if (ABS (E2) >=abs (E3))

F11=f2

E11=c1

F22=f1

E12=b1

F33=f3

E13=d1

Else

F11=f3

E11=d1

F22=f1

E12=b1

F33=f2

E13=c1

End

E1=-f22/f11*e11+e12

F1=-f33/f11*e11+e13

P1=e1 (ON)

Q1=F1 (ON)

if (ABS (p1) >=abs (Q1))

A1=p1

A2=q1

Ff1=e1

Ee1=f1

Else

A1=q1

A2=p1

Ff1=f1

Ee1=e1

End

G1=-a2/a1*ff1+ee1

H1=[e11; FF1; G1]

J1=H1 (1:3,1:3)

B1=H1 (1:3,4)

X1=j1\b1

The output results are as follows:

A =

3.0100 6.0300 1.9900

1.2700 4.1600-1.2300

0.9870-4.8100 9.3400

A1 =

3.0100 6.0300 1.9900 1.0000

1.2700 4.1600-1.2300 1.0000

0.9870-4.8100 9.3400 1.0000

B1 =

3.0100 6.0300 1.9900 1.0000

C1 =

1.2700 4.1600-1.2300 1.0000

D1 =

0.9870-4.8100 9.3400 1.0000

F1 =

3.0100

F2 =

1.2700

F3 =

0.9870

F11 =

3.0100

E11 =

3.0100 6.0300 1.9900 1.0000

F22 =

1.2700

E12 =

1.2700 4.1600-1.2300 1.0000

E33 =

0.9870

E13 =

0.9870-4.8100 9.3400 1.0000

E11 =

1.2700

E11 =

1.2700 4.1600-1.2300 1.0000

E22 =

3.0100

E12 =

3.0100 6.0300 1.9900 1.0000

E33 =

0.9870

E13 =

0.9870-4.8100 9.3400 1.0000

E1 =

0-3.8295 4.9052-1.3701

F1 =

0-8.0430 10.2959 0.2228

P1 =

-3.8295

Q1 =

-8.0430

A1 =

-8.0430

A2 =

-3.8295

FF1 =

0-8.0430 10.2959 0.2228

EE1 =

0-3.8295 4.9052-1.3701

G1 =

0 0 0.0030-1.4762

H1 =

1.2700 4.1600-1.2300 1.0000

0-8.0430 10.2959 0.2228

0 0 0.0030-1.4762

J1 =

1.2700 4.1600-1.2300

0-8.0430 10.2959

0 0 0.0030

B1 =

1.0000

0.2228

-1.4762

X1 =

1592.6

-631.9

-493.6

It can be seen that: the difference between the two code is that each step of the column in the main element of the main element of the judgment above, the first direct view, so the use of specific numbers instead of, and the second pair of variables have a general definition and assignment, through the IF statement size judgment, such a method will be more general, Based on the proficiency of MATLAB.

Using the column principal element elimination method to solve the ax=b of the equations respectively, and realize with MATLAB program (most effective version)