How can we get the basic solution system of linear algebra? How about alpha?

How can we get the basic solution system of linear algebra? How about alpha?

Let x = (a, B, c)
Then 2A + 5B = 0
Take a as any non-zero number to get a = 1, B = - 0.4
Then we take in the equation a-2b-c = 0 to get C
In this way, we can get a solution (a, B, c), and the basic solution system will come out

How to find the basic solution system? In linear algebra

Look at the line substitute book, first find the eigenvalue, then find the eigenvector corresponding to the eigenvalue. The line combination of all eigenvectors is the basic solution system

Basic solution system of linear algebra
Let the rank of n-order square matrix A = [AIJ] be n
η n (it's n, not r, wrong number above) = [an1, an2 Ann] t is a basic solution system of the system of equations (I), where AIJ is an algebraic cofactor of element AIJ in determinant | a |.

A is invertible, so we know that a * is invertible by AA * = det (a) e. therefore, the N-R vectors given in the title are the last N-R sequences of a * and are linearly independent, as long as we prove that they are the solutions of the first system of equations. As we know by AA * = det (a) e, the multiplication of the I (I = 1,2.., R) row of a and the J (J = R + 1,..., n) column of a * is 0, which just means that they are the solutions of (1)