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:
1 17 duhhhhhh
Step-by-step explanation:
Answer:197.8
The formula for a cylinder is 3.14*r^2*h but if you don't round to the nearest tenth the the answer is 197.82034.
Step-by-step explanation:Step one mutiply 3^2=9, Step two mutiply 3.14*
9=28.26, Step three mutiply 28.26*7=197.82.But if you round 197.82034 to the nearest tenth which is moving one decimal place to the right you'll will get 197.8.