Answer:
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.
Answer:
14
Step-by-step explanation:
Triangle is a 45 45 90 right triangle
Ratio of leg : hypotenuse = x : x√2
Given leg = 7√2
so hypotenuse x = 7√2 * √2 = 14
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