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).
:) Well, I dunno about anyone else but this question makes my head spin!
He can fill 6 boxes. 48 / 7 = 6R6.
I interpreted the remainder by removing the remainder to get the answer.
The answer is 15%
Explanation:
Team A has 42 members and team b has 18 more than team A so team b has 60 members. In order for the teams to be equal, there’d need to be 51 people on each team(60+42/2=51). There are 9 extra people on team b, so we need 9/60 , divide and you get 15%.