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.
The answer is was he number is being divided by 10
11. y = -23x - 21
You can get this by starting with y = mx + b (slope intercept form). Then put in all the knowns and solve for the b.
2 = -23(-1) + b
2 = 23 + b
-21 = b
Then add that to the end of the equation with m = -23
12. -5
The y-intercept of an equation is always the number added on at the end of an equation. It is also the number with no x attached to it.
13. 8x^9y^6
When you use the law of exponents, you need to make sure the exponent goes to each individual term. When we cube the 2, it becomes 8. When you cube x^3, you get x^3*x^3*x^3 or x^9. And with y^2 you get y^6
Answer:
-5^ -1 = 1/(-5) = -1/5
-1/y^(-4) = -(1/(1/y^4)) = -y^4
5^4 x 5^7 = 5^(4 + 7) = 5^11
Hope this helps!
:)
I think it might be a ? it sounds the most reasonable .
