An elementary ntimesn row replacement matrix is the same as the ntimesn identity matrix with at least one of the 1's replaced with some number k. This means it is a triangular matrix, and so its determinant is the product of its diagonal entries. Thus, the determinant of an elementary row replacement matrix is __________

options are ;

1.exactly one, all , at least one

2. 1's or 0's

3.Identity matrix,invertible matrix , triangular matrix or a zero matrix

4. product or sum

5. a number

Question

contestada

Respuesta :

tatendagota tatendagota · Answer 1 · 2020-03-08T11:02:01+01:00

Answer:

1's or 0's

Step-by-step explanation:

Thinking process:

The matrix property: det (AB) = det (A) det (B)

Adding the multiple of one row to another will be equivalent to left multiplication by an elementary matrix.

For example, let E be some form of matrix such that:

n x n matrix, and so E is an n x n elementary matrix which acts as an operator which adds l copies to the i row and to row j.

Applying the same row operation to B results in the matrix AB.

This, the matrix, without the loss of generality becomes:

[tex]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right][/tex]

therefore, as seen from the matrix, the triangular matrix of the product is diagonal.

The matrix will have a diagonal so the determint of A, det A = 1

thus: det (AB) = det (A)det(B) = 1 det (B) = det (B)