Question

An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with _______ exactly one at least one all of the ______ 0's 1's replaced with some number k. This means it is _______ an invertible matrix, a zero matrix, a singular matrix, a triangular matrix, an identity matrix, and so its determinant is the _______ sum product of its diagonal entries.​ Thus, the determinant of an elementary scaling matrix with k on the diagonal is nothing.

Answer 1

An n x n scaling matrix with k on the diagonal has k replacing all the 1's of the identity matrix, making it invertible with a determinant of k^n.

Step-by-step explanation:

An elementary n x n scaling matrix with k on the diagonal is the same as the n x n identity matrix with all of the 1's replaced with some number k. This means it is an invertible matrix, and so its determinant is the product of its diagonal entries. Therefore, the determinant of an elementary scaling matrix with k on the diagonal is k^n, assuming the matrix is of size n x n.

Answer 2

exactly one, 0's, triangular matrix, product and 1.

Step-by-step explanation:

So, let us first fill in the gap in the question below. Note that the capitalized words are the words to be filled in the gap and the ones in brackets too.

"An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with EXACTLY ONE of the (0's) replaced with some number k. This means it is TRIANGULAR MATRIX, and so its determinant is the PRODUCT of its diagonal entries.​ Thus, the determinant of an elementary scaling matrix with k on the diagonal is (1).

Here, one of the zeros in the identity matrix will surely be replaced by one. That is to say, the determinants = 1 × 1 × 1 => 1. Thus, it is a a triangular matrix.