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).
U count how many students were walking then multiply it by half itself
Answer:
The answer is 49
Step-by-step explanation:
The perimeter is the sum of all four sides so 28 divided by 4 would be 7 and 7 x 7 = 49
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Answer:
x= 0
Step-by-step explanation:
Let's first simplify the equation to make the PEMDAS process easier.
8x-10=3x-10+7x
8x-10=10x-10
Now lets start the subtracting and dividing process.
8x=10x
0=2x
0=x
Answer:
3 points
Step-by-step explanation:
in a function, an x-value must have only one y-value correlated with it.
So, you would need to remove either (-3, 2) or (-3, 3)
And, you would need to remove either (2, -2) or (2, -4)
<em>And, </em>you would need to remove either (1, 4) or (1, -4)
So--you need to remove 3 points in total for this to be a function.
[a y-value can have multiple x-values. the two points at -4 are allowed, and do not need to be removed for both of them to be there]