Methods for solving systems of linear equations, what are the methods for solving systems of linear

Clem's Law.

There are two prerequisites for solving a system of equations using Clem's rule, one is that the number of equations is equal to the number of unknowns, and the other is the determinant of the coefficient matrix.

Or it is not equal to zero. Solving a system of equations with Clem's rule is actually equivalent to using an inverse matrix.

It establishes the relationship between the solution of the system of linear equations and its coefficients and constants, but because the solution needs to calculate n+1 n-order determinants, the workload is often large, so the Clem rule is often used for theoretical proof, and rarely for specific solutions.

Matrix elimination. The augmented matrix of the system of linear equations.

The linear equations with the linear simplified stepped matrix as the augmented matrix are solved with the original equation system by the elementary transformation of the rows. When there is a solution to the system of equations, the solution of the system of linear equations can be found by taking the unknowns corresponding to the unit column vectors as non-free unknowns, and the remaining unknowns as free unknowns.

In junior high school, I have substituted the elimination method and the addition and subtraction elimination method, and the research object of the two elimination methods is placed on the equation. Taking the ternary equation system as an example, the substitution method or addition and subtraction method is used to simplify the three equations into two equations (including 2 unknowns), and then simplify the two equations into one equation (including 1 unknown quantity), first find the solution of the equation with one unknown quantity, and then find the second unknown quantity, the third unknown quantity, and so on. However, the research object of Gaussian elimination method is on the augmentation matrix, only the coefficients and constant terms of the system of equations are studied, and the three methods of equivalent transformation of the system of equations are used, such as the line wrapping method, the multiple method, and the doubling addition method, to transform the augmented matrix into [the simplest line], that is, to find the value of each unknown quantity, and the Gaussian elimination method (the equivalent exchange of the system of equations for the elementary row transformation of the augmented matrix) has universal applicability to solve the system of higher-order linear equations.

1. The concept of a system of linear equations.

1. Generally speaking, the system of linear equations we call is generally composed of unknowns (one-time), coefficients, equal signs, etc., as follows:

2. A system of linear equations can be converted into matrix form as follows:

3. Add the right end of the equation to the matrix to form an augmented matrix, which can effectively find the solution of the linear equation, as follows:

2. General solution of the system of equations.

1. The system of equations can also be written in the form of vectors as follows:

2. The concept of general solution of equations:

3. The basic methods for finding the general solution of the system of equations generally include transposition transformation, multiplication transformation, multiplication transformation, etc., as follows:

3. Row ladder equations.

1. Use the elementary row transformation to solve the following system of equations:

2. Simplification into a system of row ladder equations:

3. The concept of a system of stepped equations, as shown in the figure below.

Fourth, the classic example problem - to find a general solution.

1. Solve the general solution of the equation system of the next problem:

2. Convert to, line the system of stepped equations, and define the free unknown Qing state, the source of reputation and destruction, therefore, the general solution of the problem can be obtained, as follows:

1.Column principal element elimination method (Gaussian elimination method): This is a basic method for solving systems of linear equations.

Its idea is to transform a system of equations into an upper triangular form by eliminating elements from a matrix of coefficients. In simple terms, it reduces the system of equations to a simplified upper trigonometric form through a series of row transformation operations, and then solves for the value of the unknown number by algebra. 2.

Inverse and inverse matrix methods for matrices: If the coefficient matrix a is invertible (non-singular), the inverse matrix method can be used to solve the system of linear equations. Specifically, by writing the system of equations as a matrix ax = b, and then multiplying both sides of the equation by the inverse matrix of a at the same time, we get x = a (-1)b, and the value of the unknown vector x can be obtained.

However, note that the premise for the existence of the inverse matrix is that the coefficient matrix a is reversible. 3.Determinants of matrices and Kramer's law:

This is a method of solving systems of linear equations based on determinants and Kramer's rule. For a system of linear equations with n unknowns, the solution of the system of equations can be obtained by finding the determinant d of the coefficient matrix and the algebraic remainder of each uneliminated known number. The specific calculation steps are to first solve the determinant d of the coefficient matrix, and then replace each column of the coefficient matrix with a constant series b in turn to obtain the corresponding n algebraic remainders, and calculate the value of the unknown number through the Kramer rule after the most difficult thing.

Let x1=a,x2=b,x3=c,then 3a+2b+2c=1,a+b+2c=2,a+b+c=3, first let a+b=2-2c substitute the third, it becomes 2-2c+c=3, you can calculate c=-1, then you can simplify the equation to 3a+2b=3, a+b=4, and then let a=4-b substitute the first one, it becomes 3(4-b)+2b=3, you can calculate b=9, then you can simplify the equation again, and immediately get the last value a=- 5, that is, x1=-5, x2=9, x3=-1

The addition of Eq. 2 and Eq. 3 gives the subtraction of 5x1+4x2+x4=2 from Eq. 1 to obtain x4=1

Respectively substitute into formula 2 and take hole formula 4:

x1-x2+2x3=0

x1+x2+x3=-1

Summing: 2x1+3x3=-1

Substituting x4=1 into Eq. 1 yields: 5x1+4x3=1

The two equations form an equation and solve the friend split: x1 = 1, x3 = -1, so x2 = -1

That is: x1=1

x2=-1x3=-1

x4=1