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:
$10,234.31
Step-by-step explanation:
A suitable financial calculator or spreadsheet can evaluate the future value function for you. It will tell you that $10,234.31 must be deposited today to have $13,000 in three years, when interest is 8% compounded monthly.
__
You are solving for P:
13000 = P(1 +0.08/12)^(12×3)
P = 13000/(1 +0.08/12)^36 ≈ 10,234.31
Im sorry I cant help im in 6th grade
Answer:
The answer is C, 3 to 2
Step-by-step explanation:
This is because 12 divided by 4 is 3 and 8 divided by 4 is 2 so yhe ratio is 3 to 2. Hope it helps.
Answer:
help is for the weak as said my I, the one true burnt chicken nugget
Step-by-step explanation: