Matrix A = first row 1,1, second row 1,1 times matrix B (2) Can't I take it? B to a?

Matrix A = first row 1,1, second row 1,1 times matrix B (2) Can't I take it? B to a?

That's true,
Matrix multiplication can only be performed when the number of columns of the previous matrix is equal to the number of rows of the following matrix

Let matrix A = 1,0 in the first row and 2,1 in the second row
AX= X11 X12
2X11+X21 2X12+X22
XA= X11+2X12 X12
X21+2X22 X22
From AX = Xa, we can deduce that X12 = 0, X11 = X22, and X11, X21 can take any value
X=X11 0
X21 X11
Q: 1. How is the value of X calculated
2. The calculation process of X
3, why is X12 equal to 0
4. Why is X11 equal to X22,
5, X11, X21 why can be arbitrary value
Let a = 1 0
2 1, find out all the commutative matrices with a

Let X be a commutative matrix with a, that is, ax = Xa, because a is a 2 * 2 matrix, so x is also a 2 * 2 matrix (obtained from the restriction of order when a and X can be multiplied), so we can set X = (X11 x12x21 X22) so AX = X11 x122x

Using the elementary row transformation method to find the inverse of matrix A = the first row 123, the second row-1-24, the third row 022

(1 2 3 1 0 0 -1 -2 4 0 1 0 0 2 2 0 0 1)~(1 2 3 1 0 0 0 0 7 1 1 0 0 1 1 0 0 1/2)~(1 2 3 1 0 0 0 1 1 0 0 1/2 0 0 7 1 1 ...

The first line of the matrix is 3296, the second line is-1-36-5, and the third line is 14-73

r1-3r3,r2+r3
0 -10 30 -3
0 0 -1 -2
1 4 -7 3
So the rank of the matrix is 3, so there is
-->
1 0 0 0
0 1 0 0
0 0 1 0

Let a = (the first line 4-52, the second line 5-73, the third line 6-94), then ()
A.（1,1,1）T B.（1,1,3）T
C.（1,1,0）T D.（1,0,-3）T

4 -5 2
5 -7 3
6 -9 4
OK, and both are 1
So (a) is correct

Let a = (the first line 1 1, the second line 1 2 1, the third line 2 3 x) be 2, then x =?

Because the rank of a square matrix of order 3 is 2
So the determinant of a is equal to 0
And | a | = K-2
So k = 2

1. Let the first row of a matrix be 10 - 1, the second row be 130, and the third row be 0 21. Let X be a matrix of third order and satisfy the matrix equation AX + I = a ^ 2 + X, then find the matrix X
2. How can a negative sign be transformed into the standard form of a matrix? For example, the first line 1-10, the second line 0 1-1, and the third line 0 01
3. How to standardize the first line that is not 1? For example, the first line 2 2 3, the second line 0 - 2 - 3 / 2, the third line 0 01 / 4

1.
From ax + I = a ^ 2 + X, (A-I) x = a ^ 2-I = (A-I) (a + I)
Because A-I=
0 0 -1
1 2 0
0 2 0
Invertible (determinant = - 2)
So x = a + I=
2 0 -1
1 4 0
0 2 2
two
1 -1 0
0 1 -1
0 0 1
R2 + R3, R1 + R2 are transformed into
1 0 0
0 1 0
0 0 1
three
2 2 3
0 -2 -3/2
0 0 1/4
r3*4
2 2 3
0 -2 -3/2
0 0 1
r1-3r3,r2+3/2r3
2 2 0
0 -2 0
0 0 1
r1*(1/2),r2*(-1/2)
1 1 0
0 1 0
0 0 1
r1-r2
1 0 0
0 1 0
0 0 1

The first row of the matrix is 000, the second row is 100, and the third row is 010. Try to determine all the matrices that are commutative with a multiplication, that is, the matrix X that satisfies the condition AX = Xa
Please save the great God!

If you write out the nine elements of X, you can see it by multiplying it. X is a matrix in the form of upper triangle. The diagonal elements are equal and the sub diagonal elements are equal,

Find the eigenvalues of 3 * 3 matrix, the first row 2, - 2, 0; the second row - 2, 3, 2; the third row 0, 2, 4

|A - λ e | = 2 - λ - 20-23 - λ 20 24 - λ R3 + R1, C1 + C32 - λ - 200 3 - λ 26-2 λ 0 4 - λ = (2 - λ) (3 - λ) (4 - λ) - 2 * 2 * (6-2 λ) = (3 - λ) [(2 - λ) (4 - λ) - 8] = (3 - λ) (λ ^ 2-6 λ) = λ (3 - λ) (λ - 6). Therefore, the eigenvalues of a are 0,3

Find the eigenvalues and eigenvectors of the first line 2,3,2, the second line 1,8,2, the third line - 2, - 14,13 of the matrix
It's 232
one hundred and eighty-two
-2-14-3

|A-λE|=
2-λ 3 2
1 8-λ 2
-2 -14 -3-λ
= -(λ-1)(λ-3)^2=0
The eigenvalues are 1,3,3
1
(A-E)x=0
Coefficient matrix:
1 3 2
1 7 2
-2 -14 -4
The results of elementary row transformation are as follows
1 0 2
0 1 0
0 0 0
So the eigenvector is [- 20 1] ^ t
3 corresponding eigenvectors:
(A-3E)x=0
Coefficient matrix:
-1 3 2
1 5 2
-2 -14 -6
The results of elementary row transformation are as follows
1 1 0
0 2 1
0 0 0
So the eigenvector is [1 - 1, 2] ^ t