How to express a matrix as the product of Elementary Matrices

Premise a is reversible!
A is transformed into identity matrix by elementary row transformation, and the elementary transformation is recorded every time
This is equivalent to multiplying a series of corresponding Elementary Matrices on the left side of A
That is, PS... P1A = E
So a = P1 ^ - 1... PS ^ - 1
Because Pi is an elementary matrix, PI ^ - 1 is also an elementary matrix
So a is the product of elementary matrices

The following invertible matrices are expressed as the product of elementary matrices
1 -1
1 1
How did you do it

1 -1 1 01 1 0 1→2 0 1 11 1 0 1.2 0 1 01 1 0 1→1 0 1/2 0 1 1 0 11 0 1 0 1 1 0 1→1 0 1 0 1 1 -1 1→P1=1 1 0 1P2=1/2 1 0 1P3= 1 0 -1 11 -1 1 -1 2 0 1 0 A=1 1 = 0 1 × 0 1 × 1 1

Why can a matrix be expressed as the product of Elementary Matrices, then a must be reversible

1. Elementary matrix must be invertible (and the inverse matrix is also elementary matrix)
2. The product of finite invertible matrices must be invertible, and (P1... PK) ^ {- 1} = PK ^ {- 1}... P1 ^ {- 1}
These are the most basic conclusions. Take a good look at the textbook and take your time

1. The square matrix AB (a is 3 * 2, B is 2 * 3) must be irreversible. 2. The product of two n-order Elementary Matrices must be reversible. Why is 3, a a a third-order square matrix
1. The square matrix AB (a = 3 * 2, B = 2 * 3) must be irreversible
2. The product of two n-order Elementary Matrices must be invertible. Why
3. If a is a third-order square matrix, then a is not equal to 0, the square of a is not equal to 0, | a | = 0, why?

1. The rank of matrix AB is R (AB) ≤ min {R (a), R (b)} ≤ 2, a is 3 * 2, B is 2 * 3, and their maximum rank is 2. The necessary and sufficient condition for a third-order matrix to be reversible is that R (AB) = 3, so AB must be irreversible
2. The elementary matrix is the matrix obtained by an elementary transformation of the identity matrix I (some versions are e, in short, the identity matrix). Let these two n-order Elementary Matrices be E1 and E2, then according to the properties of the elementary matrix, there must be n-order invertible square matrices P1, Q1; P2, Q2, such that E1 = P1 · I · Q1, E2 = P2 · I · Q2. (this property should be found in the book, in the elementary transformation). So E1E2 = P1 · Q1 · P2 · Q2. P1, Q1, P2, Q2 is an invertible square matrix of order n, so E1E2 is an invertible square matrix of order n
Third, I don't quite understand the meaning of the title
If "a is a third-order square matrix, if the square of a is not equal to 0, | a | = 0, then a is not equal to 0." this is true. The rank of a third-order square matrix is R (a) ≥ R (the square of a) (the property of rank), and the square of a is not equal to 0, then R (the square of a) ≥ 1, so r (a) ≥ 1, so a is not equal to 0 (the necessary and sufficient condition of a zero matrix is that the rank is equal to 0)
If "if a is a third-order square matrix, then a is not equal to 0, the square of a is not equal to 0, | a | = 0", obviously, if a is a third-order square matrix, it is impossible to deduce that a is not equal to 0, the square of a is not equal to 0, | a | = 0, such as the third-order unit matrix
If "a is a third-order square matrix, if a is not equal to 0 and the square of a is not equal to 0, then | a | = 0" is also wrong. The counterexample can still be a third-order unit matrix
If "a is a third-order square matrix, if the square of a is not equal to 0, then | a | = 0, and a is not equal to 0," this is not true. The counterexample can still be a third-order unit matrix
Wordy so much, final exam refueling ah!
If you think it's good, take it as the best answer

How to calculate the product of matrix

Cij=ai1bij+ai2b2j+...+ainbnj

Calculate the product of two 5 * 5 matrices

Because you didn't say the specific formula, so you can only provide the properties of determinant. With this, it's easy to calculate determinant. Property 1 determinant is equal to its transposed determinant

Calculating the product of matrices
1 0 0 x1 y1 z1
A= 0 1 2 B= x2 y2 z2
0 1 - 2 X3 Y3 Z3 find a * b =? B * a =?

A*B=
x1 y1 z1
x2+2x3 y2+2y3 z2+2z3
x2-2x3 y2-2y3 z2-2z3
B*A=
x1 y1+z1 2y1-2z1
x2 y2+z2 2y2-2z2
x3 y3+z3 2y3-2z3

Is the product of elementary matrix still elementary matrix? The answer given in the book is wrong. Can you give a counter example?

Any invertible matrix can be written as the product of elementary matrix, so this conclusion is wrong
matrix
1 1
0 1
The matrix is elementary
1 0
2 1
It is also an elementary matrix
3 1
2 1
The matrix can not be obtained by an elementary transformation of the identity matrix
The product of elementary matrix is invertible matrix

Try to express a as the product of Two Elementary Matrices and find the process
A=1 0 0
1 1 0
0 1 1

First of all, we should know that the elementary row transformation of matrix A is equivalent to multiplying a by the corresponding elementary matrix to the left. Similarly, the elementary column transformation of matrix A is equivalent to multiplying a by the corresponding elementary matrix to the right. Consider the matrix in the question, change a slightly, so that B = 100
1 1 0
0 0 1
This is an elementary matrix (obtained by adding the first row of the identity matrix to the second row). Now if the third column of B is added to the second column, a will be obtained. Therefore, it is equivalent to the right multiplication of matrix B by an elementary matrix C, and C is obtained by adding the third column of the identity matrix to the second column
C=1 0 0
0 1 0
0 1 1
So a = BC

How to find the product C (mxn) of two matrices A (m × n), B (kxn)

It must be k = n

How to express a matrix as the product of Elementary Matrices