Answer:
The value of AB is
and it's not possible to multiply BA.
Step-by-step explanation:
Consider the provided matrices.
, ![B=\left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.
Multiply AB
Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:
![AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5Ctimes%203%2B3%5Ctimes%205%5C%5C2%5Ctimes%203%2B1%5Ctimes%205%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%2B15%5C%5C6%2B5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Hence, the value of AB is ![\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Now calculate the value of BA as shown:
Multiply BA
Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.
We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.
That means number of column of first matrix should be equal to the number of rows of second matrix.
Hence, it's not possible to multiply BA.
The nine rounds up so it would be 400,000
His first payment is $100, thus a₁ = 100.
the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.


is the serie divergent or convergent?
well, to make it short, when the common ratio is 0 < | r | < 1, namely a fraction between 0 and 1, only then the serie is convergent, namely it reaches a fixed value, now in this case, 1.1 is a value larger than anything between 0 and 1, so no dice.
Answer: imma figure this out
Step-by-step explanation:
Answer:
non living are computer baseball phone bear living are tree human fern and bacteria the rest i.d.k
Step-by-step explanation: