Question

You are choosing between two monthly long-distance phone plans offered by two different companies. Company A charges $$\$ 0.05$$ per minute for in-state calls, $$\$ 0.12$$ per minute for state-to-state calls, and $$\$ 0.30$$ per minute for international calls. Company \(\mathrm{B}\) charges $$\$ 0.085$$ per minute for in-state calls, $$\$ 0.10$$ per minute for state-to-state calls, and $$\$ 0.25$$ per minute for international calls. In a month, you normally use 20 minutes on in-state calls, 60 minutes on state-to-state calls, and 30 minutes on international calls. (a) Write a matrix \(C\) that represents the charges for each type of call \(i\) by each company \(j\). State what each entry \(c_{i j}\) of the matrix represents. (b) Write a matrix \(T\) that represents the times spent on the phone for each type of call. State what each entry of the matrix represents. (c) Find the product \(T C\) and state what each entry of the matrix represents. (d) Which company should you choose? Explain.

Short answer

The matrix representing call charges is \(C = \begin{bmatrix} 0.05 & 0.12 & 0.30\ 0.085 & 0.10 & 0.25\ \end{bmatrix}\) and the time matrix is \(T = \begin{bmatrix} 20 & 60 & 30\ \end{bmatrix}\). The matrix product \(T \cdot C = \begin{bmatrix} 4.55 & 7.20\\end{bmatrix}\) represents the total cost for company A and B respectively. Therefore, Company A is the better choice because it costs less.

Step-by-step solution

Create the charge matrix

Create a 2x3 matrix \(C\) for the charges. Rows represent companies (A and B) and columns represent types of calls (in-state, state-to-state, international), i.e. \[ C = \begin{bmatrix} 0.05 & 0.12 & 0.30\ 0.085 & 0.10 & 0.25\ \end{bmatrix} \] Each entry \(c_{ij}\) of this matrix represents charges per minute for the respective company and call type.

Create the time matrix

Create a 1x3 matrix \(T\) for the times. The columns represent types of calls (in-state, state-to-state, international), i.e.\[ T = \begin{bmatrix} 20 & 60 & 30\ \end{bmatrix} \] Each entry \(t_{i}\) of this matrix represents the time spent in minutes for the respective call type.

Find the product \(TC\)

Multiply the matrices. Each entry of the product matrix represents the total cost for the respective company, i.e.\[ T \cdot C = \begin{bmatrix}20 & 60 & 30\\end{bmatrix} \cdot \begin{bmatrix} 0.05 & 0.12 & 0.30\ 0.085 & 0.10 & 0.25\\end{bmatrix} = \begin{bmatrix} 4.55 & 7.20\\end{bmatrix} \]The first entry represents the total cost for company A and the second entry represents the total cost for company B.

Choose the company

Select the company that costs less based on the entries of the product matrix. Company A costs \(4.55\) units and Company B costs \(7.20\) units. Therefore, choosing Company A will be cheaper.

Key concepts

Cost Analysis in Algebra

Analyzing costs through algebra helps determine the most economical choice between different service providers. In this exercise, two phone companies provide various rates for different types of calls. Algebraic methods allow us to compare costs efficiently, rather than manually calculate costs for each scenario.

By constructing a cost matrix (let's call it matrix \( C \)), you can summarize the charges per minute: the rows represent Company A and Company B, while the columns stand for in-state, state-to-state, and international calls. This is a compact way to show which company charges what for each type of call. By using this structured approach, you seamlessly plug in real numbers to find which plan best suits your budget.

• Company A: \( \begin{bmatrix} 0.05 & 0.12 & 0.30 \end{bmatrix} \)
• Company B: \( \begin{bmatrix} 0.085 & 0.10 & 0.25 \end{bmatrix} \)

Breaking down complex billing into this simple format allows anyone to quickly visualize and compare expenses. It creates a straightforward pathway to decision-making by providing clear quantitative insights.

Matrix Representation of Charges

Representing charges using matrices is an efficient way to organize and calculate multiple rates. Each element of a matrix has a specific interpretation that simplifies the observation and manipulation of data.

In this example, matrix \( C \) represents charges for two companies. The matrix built as \( C = \begin{bmatrix} 0.05 & 0.12 & 0.30 \ 0.085 & 0.10 & 0.25 \end{bmatrix} \) formalizes comparison between Company A and B. Each column corresponds to a type of call, whereas each row identity matches a respective company.

• First row: Company A's rates for in-state, state-to-state, and international calls.
• Second row: Company B's rates for the same categories.

This matrix is easy to expand or modify—if rates change, you can simply alter the respective matrix elements. The matrix eliminates potential errors from more scattered computations, ensuring clarity.

Linear Algebra in Problem Solving

Linear algebra, through matrix multiplication, serves as a powerful tool to resolve practical problems such as determining phone plan costs. Multiplication of a time matrix by a charge matrix yields a straightforward interpretation of data across the required dimensions.

Matrix \( T \), representing duration of calls, is \( T = \begin{bmatrix} 20 & 60 & 30 \end{bmatrix} \). When multiplied by the cost matrix \( C \), the resulting product \( T \cdot C = \begin{bmatrix} 4.55 & 7.20 \end{bmatrix} \) instantaneously provides total service costs per company. Each matrix entry embodies a logical multiplication of time spent and costs incurred:

• \( 4.55 \): Total monthly cost with Company A.
• \( 7.20 \): Total monthly cost with Company B.

Through the simplicity of calculating \( T \cdot C \), decisions become intuitive. Linear algebra doesn’t only solve equations; it directly applies mathematical concepts to everyday choices, optimizing decision-making by precisely quantifying results.