Solutions of n-ary linear equations and their criteria

1. introduce

example 1. Solve a system of linear equations :
$$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 3x_1+4x_2+2x_3=9\\ -x_1-5x_2+4x_3=10\\ 2x_1+7x_2+x_3=1\end{array}$$
Explain :
$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 3x_1+4x_2+2x_3=9\\ -x_1-5x_2+4x_3=10\\ 2x_1+7x_2+x_3=1\end{array}$$\stackrel{eliminateGo\; to2,3,4Of{x}_1}{*}$$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 0-5x_2-x_3=3\\ 0-2x_2+5x_3=12\\ 0+x_2-x_3=-3\end{array}$$\stackrel{(2,4)}{*}$$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 0+x_2-x_3=-3\\ 0-2x_2+5x_3=12\\ 0-5x_2-x_3=3\end{array}$$\stackrel{eliminateGo\; to3,4Of{x}_2}{*}$$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 0+x_2-x_3=-3\\ 0+0+3x_3=6\\ 0+0-6x_3=-12\end{array}$$\stackrel{eliminateGo\; to4Of{x}_3}{*}$$\{\begin{array}{l}{x}_1+3x_2+x_3=2\\ 0+x_2-x_3=-3\\ 0+0+3x_3=6\\ 0+0+0=0\end{array}$

analysis : The above process of solving equations , Only coefficients and constant terms are changing , and $x_1,x_2,x_3$ It only plays a role of occupying space , Therefore, we can form a matrix with constant terms and coefficients according to their invariable positions , Thus, the solution of the equation is transformed into a simpler matrix operation .

$\left(\begin{array}{cccc}1& 3& 1& 2\\ 3& 4& 2& 9\\ -1& -5& 4& 10\\ 2& 7& 1& 1\end{array}\right)$$\stackrel{eliminateGo\; to2,3,4Of{x}_1}{*}$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& -5& -1& 3\\ 0& -2& 5& 12\\ 0& 1& -1& -3\end{array}\right)$$\stackrel{(2,4)}{*}$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& 1& -1& -3\\ 0& -2& 5& 12\\ 0& -5& -1& 3\end{array}\right)$$\stackrel{eliminateGo\; to3,4Of{x}_2}{*}$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& 1& -1& -3\\ 0& 0& 3& 6\\ 0& 0& -6& -12\end{array}\right)$$\stackrel{eliminateGo\; to4Of{x}_3}{*}$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& 1& -1& -3\\ 0& 0& 3& 6\\ 0& 0& 0& 0\end{array}\right)$

A matrix consisting of coefficients and constant terms of the system of equations is called Augmented matrix . The augmented matrix in this example is as follows :
$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 3& 4& 2& 9\\ -1& -5& 4& 10\\ 2& 7& 1& 1\end{array}\right)$$
After a series of elimination , Finally, we get a matrix that looks like a ladder , It's called Ladder matrix , As shown below :
$$\left(\begin{array}{cccc}1& 3& 1& 2\\ 0& 1& -1& -3\\ 0& 0& 3& 6\\ 0& 0& 0& 0\end{array}\right)$$
At this point, the solution of the equations can be obtained . that Ladder matrix It has the following characteristics :

1. 0 That's ok ( All elements are 0) Below ;
2. Principal component ( First non 0 element ) The column index of strictly increases with the increase of row index .

Carefully implement the process of elimination method , The corresponding matrix performs the following operations :

1. Add multiples of one line to another ;
2. Two lines swap ;
3. Multiply a line by a non-zero number .

The above operation on the augmented matrix of equations is called " Elementary row transformation of matrix "( Explain later ), in other words , The equations obtained by the elementary row transformation of the matrix and the original equations are the same solution equations ! In this way, we can use matrix to solve linear equations :

1. Write the augmented matrix of the system of equations ;
2. The step matrix is obtained by elementary row transformation of the augmented matrix ;
3. According to the step matrix, the solution of the equations is obtained .

Then according to the example 1 The matrix elimination method demonstrated in solves the following equation , So that we can further understand the solution of linear equations .
example 2. Solve equations :
$$\{\begin{array}{l}{x}_1-x_2+x_3=1\\ x_1-x_2-x_3=3\\ 2x_1-2x_2-x_3=3\end{array}$$
Explain :
$\left(\begin{array}{cccc}1& -1& 1& 1\\ 1& -1& -1& 3\\ 2& -2& -1& 3\end{array}\right)$$\stackrel{eliminateGo\; to2,3Of{x}_1}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& -2& 2\\ 0& 0& -3& 1\end{array}\right)$$\stackrel{c_2\times (-\frac{1}{2})}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& 1& -1\\ 0& 0& -3& 1\end{array}\right)$$\stackrel{eliminateGo\; to3Of{x}_3}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& 1& -1\\ 0& 0& 0& -2\end{array}\right)$

Write the equations corresponding to the ladder matrix obtained above as follows :
$$\{\begin{array}{l}{x}_1-x_2+x_3=1\\ 0+0+x_3=-1\\ 0+0+0=-2\end{array}$$
In the above equations , There is an equation called : $0=d(Not0)$ . Therefore, the equations have no solution .

example 3. Solve equations :
$$\{\begin{array}{l}{x}_1-x_2+x_3=1\\ x_1-x_2-x_3=3\\ 2x_1-2x_2-x_3=5\end{array}$$
Explain :
$\left(\begin{array}{cccc}1& -1& 1& 1\\ 1& -1& -1& 3\\ 2& -2& -1& 5\end{array}\right)$$\stackrel{eliminateGo\; to2,3Of{x}_1}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& -2& 2\\ 0& 0& -3& 3\end{array}\right)$$\stackrel{c_2\times (-\frac{1}{2})}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& 1& -1\\ 0& 0& -3& 3\end{array}\right)$$\stackrel{eliminateGo\; to3Of{x}_3}{*}$$\left(\begin{array}{cccc}1& -1& 1& 1\\ 0& 0& 1& -1\\ 0& 0& 0& 0\end{array}\right)$

Write the equations corresponding to the ladder matrix obtained above as follows :
$$\{\begin{array}{l}{x}_1-x_2+x_3=1\\ 0+0+x_3=-1\\ 0+0+0=0\end{array}$$
It can be obtained that the stepped equations have infinite solutions , Thus, the original equations also have infinite solutions !

2. Summative guess

The shape is as follows n A system of elementary linear equations :
$$\{\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+...+a_{1n}{x}_n=b1\\ a_{21}{x}_1+a_{22}{x}_2+...+a_{2n}{x}_n=b2\\ ...\\ a_{n1}{x}_1+a_{n2}{x}_2+...+a_{nn}{x}_n=b_n\end{array}(1)$$
Equations (1) There are only three cases of solutions : unsolvable , There is a unique solution , There are infinite solutions .

By putting the equations (1) Of Augmented matrix Transformed into Stepped matrix . If the corresponding ladder equations appear "0=d(d It's nonzero )", Then the original equations unsolvable ; Otherwise, the original equations Have solution .
When there is a solution , If the number of non-zero rows of a stepped matrix r=n(n Is the number of unknowns ), Then the original equations have Unique solution ; if r<n, Then the original equations have Infinite solutions .

The above is about n The solutions of linear equations and their criteria need to be further proved before they can be confirmed , But first of all, it's right to tell you .