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).
The probability of hitting the black circle is closer to 0 than it is to 1. because: find area of circle ~ 3.14*r^2
which is 3.14 because the radius squared is 1. so 1*3.14= 3.14
and the area of the whiteness is 121.
subtract 3.14 from 121 and you get 117.86
this answer is closer to a 0 probability than the white space. the white space is closer to a probability of 1.
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Answer:
The correct answer is <em><u>8 2/3yards</u></em>
I don’t know, I forgot how to solve problems like this, look it up
Answer:
x(1,4) Y(2,3) and Z(5,y)
Step-by-step explanation: