Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
.
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
.
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
Answer: Jose
Step-by-step explanation: Because 2/3 is the same as 8/12
Answer:
k = - 6
Step-by-step explanation:
Given a function f(x), then a horizontal translation parallel to the x- axis is denoted by
f(x + k)
If k > 0 then shift of k units left
If k < 0 then shift of k units right
Here the shift is 6 units to the left, from 4 to - 2, hence
k = 6
The answer is 50/100 same as the decimal sort of
Answer:
Rounded to the nearest tenth they are -14.5 and 12.5
Step-by-step explanation:
The zeros of a function are the x-intercepts or roots where the function crosses the x-axis. To find them, graph the function x^2 +2x-180 and zoom in on the x-axis.
See attached picture.