C solves a system of equations with matrices 100

Works: Calculation of a system of multivariate equations.

Date: Saturday, April 25, 2009.

Note: The output value can only be a decimal place (up to six digits), for example, x=

han: The maximum number of unknowns set at the beginning can only be smaller than the number after running the program.

hang: the number of unknowns that you change during the calculation.

juzhen: The initial example matrix.

#include

#include

define han 200 (can be set) multivariate linear equations have n rows n+1 columns (the more column is the value to the right of the equal sign), give the number of trips to determine the matrix, define juzhen ,, example one.

main()

int i,j,k,m,n,t,cf,hang=4;

float temp;

float aa[han][han+1]=;Define the array to be evaluated.

do determines whether to retry.

for(i=0;i=0;k--) This big loop converts the top right corner of the array to 0

for(i=k+1;iaa[k][hang+1-1]-=aa[k][i]*aa[i][hang+1-1];

aa[k][i]=0;

for(i=0;iprintf("");

for(j=0;jprintf("%g\t",aa[i][j]);

printf("The value of the unknown is:");

for(i=0;iprintf("x(%d)=\t%g",i+1,aa[i][hang+1-1]);

printf("");

printf("Do you want to try again? Yes:1; No: 0");

scanf("%d",&cf);

while(cf==1);Determine whether to try again.

1) From a2, 6, a3 into a series of equal differences, 12 = a2 + a3 ...(2 points) is also a proportional series, and a1=2, so 12=2q+2q2....(3 points) The solution gives q=2, or q=-3, and q 0....(5 points), q=2, an 2?2n?1＝2n…(7 points).

2）∵bn＝log22n＝n，

Summary. The explanation of the step of solving the system of equations matrix is to let the coefficient matrix of the equation be a, the unknown matrix is x, and the constant matrix is b, that is, ax=b, and x, then both ends of the equation are left multiplied a (-1) at the same time, and x=a (-1)b. And because (a,e) (e,a(-1)), a (-1) can be found with elementary row transformations, so that all unknowns are found.

The value of an unknown that equalizes the left and right sides of the equation is called the solution of the equation. The process of finding the solution of an equation is called solving an equation. An equation that must contain an equation of unknowns is called an equation.

An equation doesn't have to be an equation, an equation has to be an equation.

The step of solving the equation system matrix is to let the coefficient matrix of the equation buried clear be a, the unknown major answer matrix is x, and the constant matrix is b, that is, ax=b, and x, then both ends of the equation are left multiplied a (-1) at the same time, and x=a (-1)b. And because (a,e) (e,a(-1)), a (-1) can be found with elementary row transformations, so that all unknowns are found. The value of an unknown that equalizes the left and right sides of the equation is called the solution of the equation.

The process of finding the solution of an equation is called solving an equation. An equation that must contain an equation of unknowns is called an equation. An equation doesn't have to be an equation, an equation has to be an equation.

Find his adjoint matrix, and you can find this matrix.

How to ask for it. With its adjoint matrix, you're good to go.

2,1)(3,1) is the answer.

Yes. Can you give a detailed process?

The step of solving the equation system matrix is to let the coefficient matrix of the equation buried clear be a, the unknown major answer matrix is x, and the constant matrix is b, that is, ax=b, and x, then both ends of the equation are left multiplied a (-1) at the same time, and x=a (-1)b. And because (a,e) (e,a(-1)), a (-1) can be found with elementary row transformations, so that all unknowns are found. The value of an unknown that equalizes the left and right sides of the equation is called the solution of the equation.

The process of finding the solution of an equation is called solving an equation. An equation that must contain an equation of unknowns is called an equation. An equation doesn't have to be an equation, an equation has to be an equation.

By constructing an augmented matrix with a coefficient matrix and a constant matrix, and transforming the elementary rows into the simplest row matrix, a solution system is obtained, so that different constants are multiplied by the column vectors of the solution system, and the basic solution system is obtained.

For example: i1= (1 2,1 2)cos(2 t+ )e (-j t)dt,i2= (1 2,1 2)sin(2 t+ )e (-j t)dt

Then: i=i1+ji2= (1 2,1 2)e [j(2 t- t+ )dt=[e (j )]1 2,1 2)e [j(2 - t]dt=[e (j )]j(2 -

So: i=[e (j )]j(2 - e (j )]2j)sin( 2)] j(2 - e (j )]2sin( 2)] 2 -

So: i1=2[(cos)sin( 2)] 2 -

So: original = 2i1 = 4 [(cos )sin( 2)] 2 -

Require a =

2 1 1],b =

5],x =

a bc], then ax = b, and the elementary row transformation of [a, b] yields 0 0 1 4], from which we get.

x = a^b

i.e. a = 14, b = 19, c = 4

where the transformations are, in order.

2 rows - 3 rows 1 row - 1 3 times 2 rows 2 rows * 1 3 rows * 1 2 1 rows - 3 rows.

Solution: Augmentation matrix (a,b)=

r2-r3,r3*(1/2),r1-r3

r2*(1/3),r1-r2

r3-r1 swap line 1 0 0 -2

The system of equations has a unique solution: (x,y,z)=(-2,2,0)

The basic idea of solving a system of equations is "elimination and descent":

"Elimination" mainly eliminates the unknown through addition, subtraction, division and substitution until there is only one left; "Downgrading" is mainly achieved by division and factoring, until the number of unknowns is reduced to once.

Method 1: The matrices corresponding to the two systems of equations are transformed into trapezoidal matrices, and if they can be turned into the same trapezoidal matrix, the two systems of equations are solved in the same way.

Method 2: First find the rank of the matrix corresponding to a system of equations, form a system of equations into a system of equations, and then find the corresponding rank

The problem may be simple or it may not be solved at all, as follows:

If the equation you give is the easiest way to solve the analytic expressions of x and y by hand, and then let the computer perform the specific operation, it should not be difficult for your equation.

If you want to implement a general multivariate system solver, there are two scenarios:

There are definite methods to solve linear multivariate equations, such as Gaussian elimination method in linear algebra, QR decomposition method, etc., but the amount is very large, and there are related programs on the Internet that can be searched, but it also requires you to have a certain basic knowledge of linear algebra;

For the system of nonlinear multivariate equations, there is no definite solution in theory, and it must be converted into a system of linear equations before solving it according to the specific situation, but this transformation is not necessarily feasible, and it may not be possible to turn it out at all, for example, the system of equations you give is not possible.

If possible, it is recommended that the landlord use MATLAB to solve the problem.

For the specific solution, you can search for "MATLAB solution equations".

Augmentation matrix.

Make a line elementary transformation.

This line is unchanged. This line is line 1.

This line is line 1 3

The line 2 8 7

This line is unchanged. This line 2 is 4

Solution. x1=15u/7-3v/7+8/7

x2=-6u/7+4v/7+1/7

x3=ux4=v

Augmentation matrix. Make a line elementary transformation.

7 14 line 2 line 2

5 This line is unchanged.

28 LINE 2 3

7 14 This line 2 4

7 14 This line is unchanged.

1 LINE 1 2 7

This line is line 1 2

This line is line 1.

Solution. x1=-2t-1

x2=t+2

x3=t

x = b right multiplication (inverse of a).

Or hard, set x=x1

x2x3x4x5

x6x7x8x9

Substituting, 9 equations and 9 unknowns.

The coefficient matrix is a transformation, and the determinant of the coefficient matrix is equal to the determinant of a, and the determinant of a = 2+2+2=6 is not zero, so the system of equations has a unique solution.

a^(-1)a = e

c^(-1)c = e

In the above equation, a (-1) is the inverse matrix of a e is the identity matrix, so a (-1) * (ax + by) = a (-1) * mc (-1) * (cx + dy) = c (-1) * n so ex + a (-1) * by = a (-1) * mex + c (-1) * dy= c (-1) * n so y(a (-1) * b - c (-1) *d) = a (-1) * m - c (-1) *n

Then y = (a(-1)*m - c(-1)*n) (a(-1)*b - c(-1)*d).

Then substitute the original formula to find x.