Answer:
[tex]C_{22} = 389[/tex]
[tex]Max(A_{31}) =376[/tex]
Step-by-step explanation:
Given
See attachment for complete question
Solving (a): The entry C22
First, matrix C represents the inventory at the end of July.
The entry of C is calculated as:
[tex]C = Inventory - Sales[/tex]
i.e.
[tex]C = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right] - \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right][/tex]
[tex]C = \left[\begin{array}{ccc}543-102&356-78&643-97\\364-98&476-87&419-59\\376-54&903-89&409-79\end{array}\right][/tex]
[tex]C = \left[\begin{array}{ccc}441&278&546\\266&389&360\\322&814&330\end{array}\right][/tex]
Item C22 means the entry at the second row and the second column.
From the matrix
[tex]C_{22} = 389[/tex]
Solving (b): The maximum A31 possible.
From the given data, we have:
[tex]Inventory = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right][/tex]
[tex]Unit\ Sales = \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right][/tex]
From the matrices above.
A31 means entry at the 3rd row and 1st column.
So, the possible values of A31 are:
[tex]A_{31} = 376[/tex]
and
[tex]A_{31} = 54[/tex]
By comparison, 376 > 54
So:
[tex]Max(A_{31}) =376[/tex]
