The matrix is not properly formatted.
However, I'm able to rearrange the question as:
![\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C-2%263%265%7C3%5C%5C3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
Operations:


Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.
Answer:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The first operation:

This means that the new second row (R2) is derived by:
Multiplying the first row (R1) by 2; add this to the second row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by 2
![2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]](https://tex.z-dn.net/?f=2%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&5&7|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%265%267%7C1%5Cend%7Barray%7D%5Cright%5D)
The second operation:

This means that the new third row (R3) is derived by:
Multiplying the first row (R1) by -3; add this to the third row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by -3
![-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]](https://tex.z-dn.net/?f=-3%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Hence, the new matrix is:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Perimeter= X + Y + W. Perimeter is the sum of all sides
Two options are given to Tina. We find how much she pays for each option, and the one in which she pays less offers the better value.
Tina sends 12 text messages each day in June
June has 30 days, so in June, Tina sent 12*30 = 360 text messages.
Package A:
First 100 messages cost 3p, the next 100 cost $2 and after that each costs $1.
She sends 360 messages, with:
The first 100 costing 3p.
The next 100 costing 2p.
The final 360 - 200 = 160 costing 1p.
She pays:

Package B:
2p for each message, 360 messages, so:
360*2p = 720p.
Which package is better?
Package A costs less, thus, it offers her the better value for money in 30 days.
Another example of a problem in which a person has to choose between two packages is given in brainly.com/question/10693932
Answer:
IT would be 13a to 17b
Step-by-step explanation:
I had this.
Fifi gets 1 cup.
And Honey's weight is 50 pounds.
Hope this helped!