Mr. Liu, I want to ask the number of linearly independent vectors = rank of vector group = rank of matrix. Is that right

Right
The first one should be the number of vectors contained in the maximally independent group

The proof of the rank of matrix and the rank of its column vector group
When we prove that the rank of matrix is equal to the rank of row vector in the fourth edition of Tongji linear algebra, the process is as follows:
Proof: let a = (A1, A2,. AM) r (a) = R
Let Dr not be equal to 0, then we know that the column in which Dr is located is linearly independent by Dr not being equal to 0. Moreover, we know that any R + 1 column vector in a is linearly related by the fact that any R + 1 subexpression of order r is zero
How does it get that any R + 1 column in a is linearly related by the fact that all R + 1 subformulas are zero? I think that if all R + 1 subformulas are zero, we can only get that the matrix composed of the elements of the R + 1 determinant is linearly related
In other words, the original vector is linearly related, so after increasing the dimension, it is not necessarily linearly related

Row rank = column rank = rank of matrix

Hello, Miss Liu. If (a is m * n matrix) AX = B has infinite solutions, then the rank of the solution vector is N-R (a) + 1
On the rank of solution vector

That's right

Solve the matrix A = [0 01 2 10 - 1 1 - 1], B is a third order square matrix, and ab = 0, then the rank of matrix B is
Let a = [0 01]
2 1 0
-If B is a third order square matrix and ab = 0, then the rank of matrix B is

Because AB = 0, R (a) + R (b) = 3
And because R (a) = 3, R (b) = 0

Finding the rank of a matrix in the first column of 1 2 1 3, the second column of 4-1-5-6 and the third column of 1-3-4-1

1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3
4 -1 -5 -6 = 0 -9 -9 -18 = 0 1 1 2 = 0 1 1 2
1 -3 -4 -1 0 -5 -5 -4 0 1 1 4/5 0 0 0 1
The rank of a matrix = 3

Calculate the rank of the following matrix A = the first row 0 1 - 1 2, the second row 0 2 - 2 2 0, the third row 0 - 1 1 1 and the fourth row 1 0 1

r2-2r1,r3+r1
0 1 -1 -1 2
0 0 0 0 -4
0 0 0 0 3
1 1 0 1 -1
r2+(4/3)r3
0 1 -1 -1 2
0 0 0 0 0
0 0 0 0 3
1 1 0 1 -1
Exchange bank
1 1 0 1 -1
0 1 -1 -1 2
0 0 0 0 3
0 0 0 0 0
This is a ladder matrix, and the number of nonzero rows is the rank of the matrix
So r (a) = 3

Find the rank of the following matrix: the first row 0.11-1.2, the second row 0.2-2.0, the third row 0.1-1.1, and the fourth row 1.01-1

A=
[0 1 1 -1 2]
[0 2 -2 -2 0]
[0 -1 -1 1 1]
[1 1 0 1 -1]
Elementary transformation to
[1 1 0 1 -1]
[0
1 1 -1 2]
[0 0 -4 0 -4]
[0 0 0 0 3]
Elementary transformation to
[1 1 0 1 -1]
[0
1 1 -1 2]
[0 0 1 0 1]
[0 0 0 0 1]
Rank r (a) = 4

Let the rank of matrix A (the first row: 1, a, a, a; the second row: A, 1, a, a; the third row: A, a, 1, a; the fourth row: A, a, a, 1) be 3, then a=

Because R (a) = 3
So | a | = 0
And | a | = (3a + 1) (1-A) ^ 3
So a = 1 or a = - 1 / 3
When a = 1, R (a) = 1 does not conform to rounding
So a = - 1 / 3

Use row elementary transformation to find the rank a of the following matrix = {the first row 1 1 - 1, the second row 3 10, the third row 4 41, the fourth row 1 - 2 1}

1 1 - 13 1 04 4 1 the second line subtracts three times the first line, the third line subtracts four times the first line, the fourth line subtracts the third line to get 1 - 21 1 - 10 - 2 300 5, the third line divides by 5 to get 0 - 3 21 1 - 10 - 2 3 the second line

Matrix A = the rank of the first row 3-1-42-2, the second row 10-110, the third row 121-34 and the fourth row 143-30 is equal to?

A=3 -1 -4 2 -2
1 0 -1 1 0
1 2 1 3 4
-1 4 3 - 3 0 line 1 minus line 2 multiplied by 3, line 3 minus line 2, line 4 plus line 2
=0 -1 -1 -1 -2
1 0 -1 1 0
0 2 2 2 4
0 4 2 - 2 0 line 3 plus line 1 multiplied by 2, line 4 plus line 1 multiplied by 4
=0 -1 -1 -1 -2
1 0 -1 1 0
0 0 0 0 0
0 0 - 2 - 6 - 8 lines 1 and 2 swap, lines 3 and 4 swap
=1 0 -1 1 0
0 -1 -1 -1 -2
0 0 -2 -6 -8
Multiply line 2 by - 1 and divide line 3 by - 2
=1 0 -1 1 0
0 1 1 1 2
0 0 1 3 4
Add line 3 to line 1 and subtract line 3 from line 2
=1 0 0 4 4
0 1 0 -2 -2
0 0 1 3 4
0 0 0 0 0
Obviously, there are three non-zero lines here,
So the rank of matrix A R (a) = 3

Mr. Liu, I want to ask the number of linearly independent vectors = rank of vector group = rank of matrix. Is that right