Current location - Training Enrollment Network - Mathematics courses - In 2006, the ninth problem of mathematics for postgraduate entrance examination was non-homogeneous linear equations.
In 2006, the ninth problem of mathematics for postgraduate entrance examination was non-homogeneous linear equations.
Let me help you answer that.

1. "Yes" means that the non-homogeneous linear equations have at least three linearly independent solutions, that is, the number of linearly independent solutions is greater than or equal to three; "Exactly" means that the nonhomogeneous linear equations have only three linearly independent solutions.

2. Because α 1, α2, α3 are three linearly independent solutions of the homogeneous linear equations, it is easy to verify that α 1-α2 and α 1-α3 are two linearly independent solutions of the homogeneous linear equations, because a (α1-α 2) = α 6544. Linear independence Because there is no nonzero constant K 1, K2 makes k1(α 1-α2)+K2 (α1-α3) = 0 (otherwise α1,α 2, α 3 are linearly related), then according to the homogeneous linearity,

Number of row-by-row independent solutions =n-r(A)=4-r(A),

Because the word "you" in the title indicates the existence, there may be more than three linearly independent solutions to the nonhomogeneous linear equations Ax=β, and then it is possible to construct more than two linearly independent solutions to the corresponding homogeneous linear equations Ax=0 by imitating the method in the title. Means 4-r(A)≥2. r(A)≤2。

There is a second-order subformula in matrix A that is not zero. You can refer to the definition of matrix rank in the textbook, that is, r(A) is the highest order number of a non-zero sub-formula, but note that there is "yes" here, that is, there may be sub-formulas with more than two orders of non-zero, so the highest order number of non-zero sub-formulas with r(A)=A is ≥2.

I hope my answer can help you.