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.

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.

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.