The basic solution system has three conditions:

1 is the solution of the system of equations Ax=0.

2. This is a linear independent solution.

3. Any solution of the system of equations Ax=0 can be expressed linearly. (The implicit condition is that the number of solution vectors of the basic solution system =n-r(A))

explain

(It is proved that 1 is the solution of the system of equations Ax=0. )

α 1, α2, ..., αs is the basic solution system of the system of equations Ax=0.

α 1, α2, ..., αs can linearly represent βj, then βj is the solution of the equation set Ax=0.

(Proof: 2. This is a linearly independent solution. )

Let A=(α 1, α2, ..., αs) and B=(β 1, β2, ..., βs),

Then according to the known β 1=α2+...+αs, β2=α 1+...+αs, ......

Write in matrix form (α 1, α2, ..., αs)C = (β 1, β2, ..., βs).

Matrix c is

0 1 1 ... 1

1 0 1 ... 1

......

1 1 1 ...0

Because α 1, α2, ..., αs is the basic solution system of the system of equations Ax=0, it is linearly independent, and because |C|≠0 of matrix C is reversible.

Then (β 1, β2, ..., βs) must be linearly independent.

(3) It can represent all solutions of the equation system Ax=0. )

Because (α 1, α2, ..., αs)C = (β 1, β2, ..., βs), C is reversible.

R(A)=r(B)=s, and the number of solution vectors is equal.

n-r(A)=n-r(B)

To sum up, βj is the basic solution system of equations.

To annotate ...

The basic solution system should be considered from the above three aspects.

Newman hero 2065438+March 9, 2005 10:58:49

I hope it will help you and I hope it will be adopted.