Problems with a system of linear algebraic non homogeneous linear equations

Since r(a)=2, it is said that n=3-r(a)=1, and because a,b are its two linear independent solution vectors, the basic solution system of ax=0 is (a-b), and the general solution of this system of nonhomogeneous linear equations is k1(a-b)+a.

Because r(a)=3, it is said that n=4-r(a)=1, a(a+b)=2b, a(3b-2c)=b, so a(a+b-6b+4c)=0, that is, a+b-6b+4c is a solution of ax=0, because a+b=(2,4,6,8),3b-2c=(1,3,5,7), so the basic solution system of the second system of equations is (0,-1,-2,-3),1 2a(a+b)=b, so a set of solutions of ax=0 is (1, 2,3,4,), this general interpretation is (1,2,3,4,)+k1(0,-1,-2,-3).

Proof: The system of equations ax=b has a solution.

r(a)=r(a,b)

r(a^t)

r(a^t;

b^t)-(a^t;

b t) is the matrix of the upper and lower blocks.

b t can be used by. a t.

Row vector. Group linear representation.

a^ty=0

And. a^t;

b t)y=0.

Any solution vector y0 with a ty=0 is a solution vector with b ty=0.

It can also be thought of like this:

The system of equations ax=b has a solution.

b can be made by a.

Column vector. Group A1 ,..An linear representation.

b t can be ,.. by the row vector group a1 t of a tAn t linear representation.

Same as above.

The number of unknowns n = 4, the rank of the augmentation matrix = the rank r(a) of the coefficient matrix = 2, and the free unknowns have n-r(a) =2.

As long as each unknown quantity is not explicitly equal to a certain constant, it can be selected as a free unknown.

You can choose x3, x4, x2, x3, x2, x4.

Because it has become a stepped type, x1 is generally not selected (not impossible).

Traditionally, select from back to front, and choose n-r(a) = 2 at the end. So choose x3, x4.

ax=0 when there is no non-zero solution. then a is a full-rank matrix. Then ax=b must have a solution.

When ax=0 has an infinite number of solutions, then a must not be a full-rank matrix, and the solution of ax=b has no solution and an infinite number of solutions.

No solution: r(a)≠r(a|b)

Infinite solution: r(a) is equal to r(a|b)。And it's not a full rank.

When ax=b has no solution, we know that ax=0 must have an infinite number of solutions.

When ax=b has a unique solution, we know that a is a full-rank matrix, and ax=0 has only a zero solution.

A system of homogeneous linear equations with either a zero solution (r(a)=n) or an infinite solution (r(a) a zero solution, a non-zero unique solution. It can't happen at the same time.

If the rank of the coefficient matrix is 2, then the number of solution vectors in the basic solution system of the corresponding homogeneous linear equation system is 4-2=2, and the sum weight of the three solutions a, b, and c can be used as a basic solution system of the homogeneous linear equations, and a special solution is a 2

Therefore, the general solution is a2+k1(a-b)+k2(c-a), where k1, k2 are arbitrary constants.

First, write out the augmentation matrix and turn it into the simplest line, and the process is as follows.

x1 and x2 are step heads, so x3 and x4 are free unknowns. Let x3=t1 and x4=t2 find the general solution and express it in the form of a vector, then the basic solution system and the fixed solution can be obtained, and the process is as follows.

Proof: The system of equations ax=b has a solution.

> r(a)=r(a,b)

> r(a^t) = r(a^t; b^t)--a^t;b t) is the matrix of the upper and lower blocks.

> b t can be expressed linearly by a set of row vectors for a t<=> a ty=0 with (a t; b t)y=0 with the same solution<=> any solution vector y0 is a solution vector with ty=0.

It can also be thought of like this:

The system of equations ax=b has a solution.

>b can be ,.. by the column vector group a1 of aAn linear representation of <=>b t can be ,.. by the row vector group A1 T of a tAn t linear representation is the same as above.

ax=b has a solution=>b in the column space of a.

A ty=0 means that y is in the orthogonal complement of a's columnspace, and if b ty=0 means that b is in a's columnspace.

The two conditions are mutually reciprocal.

The calculations upstairs were clearly wrong.

It is impossible for that many scores to come up.

The elementary transformation can be done step by step.

Write out the augmentation matrix (a, b) =

4 5 3 3 -1 4 r3-r1，r4-2r2，r1-r2~1 1 1 1 1 2

0 -1 1 1 5 4 r2-2r1，r4+r3~1 1 1 1 1 2

0 0 0 0 0 0 r1-r2，r3-r2~1 0 2 2 6 6

So the general solution of the system of equations is .

C1(-2,1,1,0,0) t+c2(-2,1,0,1,0) t+c3(-6,5,0,0,1) t+(6,-4,0,0,0) t,c1c2c3 is constant.

Solving a system of homogeneous linear equations is generally to perform a primary row transformation of the coefficient matrix, and then find the general solution.

There are two commonly used solutions to solve non-homogeneous linear equations, one is to use the Kramer rule when the number of unknowns and the number of equations are equal, but it is more troublesome when there are more unknowns, and the other is to perform elementary row transformation on the augmented matrix to obtain the general solution.

Kramer's rule is not usually used to solve systems of equations, but more often to determine the solution of systems of equations. If the coefficient matrix determinant of a homogeneous linear equation is not equal to 0, then there is only a non-zero solution, and if the coefficient matrix of a non-homogeneous linear equation is not equal to 0, there is a unique solution.