Answer: Is true sometimes.
Step-by-step explanation:
I guess that here we have two matrices, A and B, that are nxn.
We can see that if those matrices can conmutate, then we can try it with some simple matrices.
![A = \left[\begin{array}{ccc}1&0\\0&-1\end{array}\right] . B = \left[\begin{array}{ccc}2&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D%20.%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Here, we would have that:
![AB = \left[\begin{array}{ccc}2&0\\-1&-1\end{array}\right]](https://tex.z-dn.net/?f=AB%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C-1%26-1%5Cend%7Barray%7D%5Cright%5D)
![BA = \left[\begin{array}{ccc}2&0\\1&-1\end{array}\right]](https://tex.z-dn.net/?f=BA%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C1%26-1%5Cend%7Barray%7D%5Cright%5D)
You can see that AB and BA are different, then the statement is not always true.
But it is true sometimes, if A or B are the identiti, then I*A = A*I, in this case would be true.
It is also true if A and B are diagonal matrices, let's prove it:
![A = \left[\begin{array}{ccc}a&0\\0&b\end{array}\right] , B = \left[\begin{array}{ccc}c&0\\0&d\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%5C%5C0%26b%5Cend%7Barray%7D%5Cright%5D%20%2C%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc%260%5C%5C0%26d%5Cend%7Barray%7D%5Cright%5D)
![AB = \left[\begin{array}{ccc}ac&0\\0&bd\end{array}\right] = BA](https://tex.z-dn.net/?f=AB%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dac%260%5C%5C0%26bd%5Cend%7Barray%7D%5Cright%5D%20%3D%20BA)
Answer:
could you posibly put a pic
Step-by-step explanation:
Answer:
(c) 1.03 per year
Step-by-step explanation:
The relationship between "growth factor" and "growth rate" is ...
growth factor = 1 + growth rate
growth factor = 1 + 3% = 1.03
Given:
Cory made 21 of the 60 baskets she attempted.
Krista made 16 out of 40 baskets she attempted.
Paul made 17 of the 50 baskets she attempted.
Sally 11 of the 55 she attempted.
To find:
Who had the greatest percentages of baskets made?
Solution:
We know that,
Using this formula, we get
From the above percentages 40% is maximum.
Therefore, Krista has greatest percentage of baskets made.
e + 1 13/16 = 2 5/16
subtract 1 13/16 from each side
e = 2 5/16 - 1 13/16
borrow from the 2
e = 1 16/16 + 5 /16 - 1 13/16
e = 1 21/16-1 13/16
e = 1 8 /16
e = 1 1/2
