The matrix row transformation is changed into a row simplified ladder matrix -1 3 2 3 3 2 5 1 1 -2 -3 1

You can calculate with any math software on the Internet
-1 3 2 3
0 11 11 10
0 0 -2 34/11

Number multiplication of matrix and elementary row transformation of matrix
In the elementary row transformation of matrix, a non-zero number is used to multiply a row of matrix, but according to the number multiplication rule of matrix, multiplying a number by a matrix should be multiplied by every element of the matrix? What's the difference

Elementary transformation is some methods of transforming the elements of a matrix, such as adding and subtracting two lines, or multiplying a certain line by a constant. The multiplication of a matrix by a number is the result of multiplying all the elements of the matrix by the constant
You may think that multiplication is very intuitive. A matrix multiplied by a number equals the following matrix. The primary change is not so simple logic relationship. It's just a way to change the appearance of the matrix. That is to say, it's not a mathematical operation, just like 5 times (x + 1) = 5x + 5. The primary transformation is like I want to change x + 1 into 5x + 1, And this change is useful in other places, such as solving equations
Maybe you haven't come into contact with the following simple example. Just like when you solve a quadratic equation of two variables, one equation is multiplied by a constant, and then the two equations are subtracted to eliminate the unknowns. The elementary transformation is the process, but the matrix is like an equation with many unknowns

Elementary row transformation of matrix
Let a be a third-order matrix, exchange the first column of a with the second column to get B, and then add the second column of B to the third column to get C, then the invertible matrix Q satisfying AQ = C is?

Q is {1 00}
0 1 1
0 0 1}

Let a be a matrix of order n over the number field P, and the number a be the n-fold eigenvalue of A. It is proved that a = AE is a quantity matrix

It is known that there exists an invertible matrix Q satisfying Q ^ - 1aq = diag (a, a,..., a) = AE
So a = q (AE) Q ^ - 1 = aQQ ^ - 1 = AE

It is proved that if any non-zero vector in P ^ n is the eigenvector of matrix A of order n over number field P, then a must be a quantity matrix

AE1 = a1e1, AE2 = a2e2,..., AEN = Anen, where a1, A2,..., an are eigenvalues, E1, E2,..., en are n columns of the unit matrix, then AE = ed, where D is a diagonal matrix with diagonal elements A1, A2,..., an. That is, a = D. considering (EI + EJ) is the eigenvalue of a, we can see that AI = AJ, that is, all diagonal elements are equal

Let a be a matrix of order n over the number field P, and the number a be the n-fold eigenvalue of A. if a is similar to a diagonal matrix on P, it is proved that a = AE is a scalar matrix

Because a can be diagonalized, the minimum polynomial of a has no double root (this is a theorem)
Since a is the n-fold eigenvalue of a, a has n elementary factors, all of which are λ - A
So the Jordan canonical form of a is diag (a, a,..., a)
So there is an invertible matrix P such that P ^ (- 1) AP = diag (a, a,..., a) = AE (this is also a theorem)
So a = paep ^ (- 1) = AE
It's over

It is proved that the multiplication of all square matrices of order n with determinant 1 over real number field is the normal subgroup of all invertible matrices of order n with respect to matrix multiplication

Let H be the set of all matrices of order n whose determinant is 1 over the field of real numbers, and the group of all invertible matrices of order n with respect to matrix multiplication be, then for any a, B ∈ h, | ab | = | a | B | = 1, | a ^ - 1 | = | a | ^ - 1 = 1, that is ab ∈ h, a ^ - 1 ∈ h, so h is a subgroup. For any a ∈ g, B ∈ g, if ab ∈ h, that is ab | = | a |

It is proved that all invertible matrices of order n on number field k are commutative matrices of order n

Go to the link below

Hello teacher, excuse me: P is the number field, a is the n-order square matrix on it, prove: there is invertible matrix B, such that (AB) ^ 2 = ab

There are invertible matrices X and Y on P such that xay = D, where D has the form diag {I, 0}, then B = YX

Let a be a real matrix M times N, and R (a) = M

The topic should be transpose a by a to m-order positive definite matrix
(AAT) t = AAT is a symmetric matrix
Take any m-dimensional vector x and investigate XT (AAT) x = ((ATX) t) ATX
Let Xi be the ith element of the vector ax, then ((ATX) t) ATX = X1 * X1 + +xn*xn>=0
R (a) = m, ATX = 0 can deduce x = 0 (because the dimension of solution space is M-M = 0)
Therefore, only when x = 0, XT (AAT) x = 0
The transpose of a times a is a positive definite matrix of order m, and the proposition is proved

The matrix row transformation is changed into a row simplified ladder matrix -1 3 2 3 3 2 5 1 1 -2 -3 1