Chapter 4.4, Problem 56STP

(a)

To determine

To list: The prices in a $4\times 3$ matrix.

Answer to Problem 56STP

The product of the matrix is not possible.

Explanation of Solution

Given:

The table list the prices at the sandwich shop is given.

Calculation:

The list of price in $4\times 3$ matrix is as follows.

  $\left[\begin{array}{ccc}4.50& 6.75& 9.50\\ 4.50& 6.75& 9.50\\ 4.00& 6.25& 8.75\\ 4.75& 7.50& 10.25\end{array}\right]$

Therefore, the $4\times 3$ matrix is $\left[\begin{array}{ccc}4.50& 6.75& 9.50\\ 4.50& 6.75& 9.50\\ 4.00& 6.25& 8.75\\ 4.75& 7.50& 10.25\end{array}\right]$ .

(b)

To determine

To list: The new prices in a $4\times 3$ matrix.

Answer to Problem 56STP

The $4\times 3$ matrix for new price is $\left[\begin{array}{ccc}0.9& 1.35& 1.9\\ 0.9& 1.35& 1.9\\ 0.8& 1.25& 1.75\\ 0.95& 1.5& 2.05\end{array}\right]$ .

Explanation of Solution

Given:

It is given that the price is cut by 20%.

Calculation:

The list of price in $4\times 3$ matrix is as follows.

  $0.20\cdot \left[\begin{array}{ccc}4.50& 6.75& 9.50\\ 4.50& 6.75& 9.50\\ 4.00& 6.25& 8.75\\ 4.75& 7.50& 10.25\end{array}\right]=\left[\begin{array}{ccc}0.9& 1.35& 1.9\\ 0.9& 1.35& 1.9\\ 0.8& 1.25& 1.75\\ 0.95& 1.5& 2.05\end{array}\right]$

Therefore, the $4\times 3$ matrix for new price is $\left[\begin{array}{ccc}0.9& 1.35& 1.9\\ 0.9& 1.35& 1.9\\ 0.8& 1.25& 1.75\\ 0.95& 1.5& 2.05\end{array}\right]$ .

(c)

To determine

To subtract: The second matrix from the first and determine the saving to the customer for each sandwich.

Answer to Problem 56STP

The subtraction is $\left[\begin{array}{ccc}-3.6& -5.4& -7.6\\ -3.6& -5.4& -7.6\\ -3.2& -5& -7\\ -3.8& -6& -8.2\end{array}\right]$ .

Explanation of Solution

Subtract the matrices $\left[\begin{array}{ccc}0.9& 1.35& 1.9\\ 0.9& 1.35& 1.9\\ 0.8& 1.25& 1.75\\ 0.95& 1.5& 2.05\end{array}\right]$ and $\left[\begin{array}{ccc}4.50& 6.75& 9.50\\ 4.50& 6.75& 9.50\\ 4.00& 6.25& 8.75\\ 4.75& 7.50& 10.25\end{array}\right]$ as follows.

  $\left[\begin{array}{ccc}0.9& 1.35& 1.9\\ 0.9& 1.35& 1.9\\ 0.8& 1.25& 1.75\\ 0.95& 1.5& 2.05\end{array}\right]-\left[\begin{array}{ccc}4.50& 6.75& 9.50\\ 4.50& 6.75& 9.50\\ 4.00& 6.25& 8.75\\ 4.75& 7.50& 10.25\end{array}\right]=\left[\begin{array}{ccc}-3.6& -5.4& -7.6\\ -3.6& -5.4& -7.6\\ -3.2& -5& -7\\ -3.8& -6& -8.2\end{array}\right]$

Therefore, the subtraction is $\left[\begin{array}{ccc}-3.6& -5.4& -7.6\\ -3.6& -5.4& -7.6\\ -3.2& -5& -7\\ -3.8& -6& -8.2\end{array}\right]$ .