Chapter 7.5, Problem 118E

(a)

To determine

To Find: To find the value of second row and first column from the given information.

Answer to Problem 118E

The answer is 6000.

Explanation of Solution

Given information:Given $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

From the given matrix

  $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

The value of $a_{21}=6000$ which represents the number of model B in warehouse 1.

(b)

To determine

To Find: To find the shipment levels after increase.

Answer to Problem 118E

The answer is $\left[\begin{array}{c}\begin{array}{ccc}5750& 4600& 9200\end{array}\\ \begin{array}{ccc}6900& 11500& 5750\end{array}\end{array}\right]$

Explanation of Solution

Given information:Given $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

Formula used: Multiplication of a matrix by scalar.

Calculation:

Since the given matrix is

  $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

When shipment level increased by 15%, then new shipments were

  $\begin{array}{l}A+15\%A\\ =\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]+15\%\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]\\ =\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]+\left[\begin{array}{c}\begin{array}{ccc}750& 600& 1200\end{array}\\ \begin{array}{ccc}900& 1500& 750\end{array}\end{array}\right]\\ =\left[\begin{array}{c}\begin{array}{ccc}5750& 4600& 9200\end{array}\\ \begin{array}{ccc}6900& 11500& 5750\end{array}\end{array}\right]\end{array}$

(c)

To determine

To Find: To find the cost of each warehouse.

Answer to Problem 118E

The cost is $8900000, $11800000 and $10100000 for three warehouses.

Explanation of Solution

Given information:Given $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

Formula used: Multiplication of two matrices.

Calculation:

Since the given matrix is

  $A=\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]$

Let matrix B represent the sale value of each model.

Therefore, $\left[\begin{array}{cc}700& 900\end{array}\right]$

Now, the cost of each warehouses can be calculated as follows

  $\begin{array}{l}BA=\left[\begin{array}{cc}700& 900\end{array}\right]\left[\begin{array}{c}\begin{array}{ccc}5000& 4000& 8000\end{array}\\ \begin{array}{ccc}6000& 10000& 5000\end{array}\end{array}\right]\\ =\left[\begin{array}{c}700\times 5000+900\times 6000\\ 700\times 4000+900\times 10000\\ 700\times 8000+900\times 5000\end{array}\right]\\ =\left[\begin{array}{c}3500000+5400000\\ 2800000+9000000\\ 5600000+4500000\end{array}\right]\\ =\left[\begin{array}{c}8900000\\ 11800000\\ 10100000\end{array}\right]\end{array}$

Hence the required cost is $8900000 for warehouse 1, $11800000 for warehouse 2 and $10100000 for warehouse 3.