Problem 49 - Algebra and Trigonometry

Question

An encyclopedia saleswoman works for a company that offers three different grades of bindings for its encyclopedias: standard, deluxe, and leather. For each set she sells, she earns a commission based on the set’s binding grade. One week she sells one standard, one deluxe, and two leather sets and makes \(\$ 675\) in commission. The next week she sells two standard, one deluxe, and one leather set for a \(\$ 600\) commission. The third week she sells one standard, two deluxe, and one leather set, earning \(\$ 625\) in commission. (a) Let \(x, y,\) and \(z\) represent the commission she earns on standard, deluxe, and leather sets, respectively. Translate the given information into a system of equations in \(x, y,\) and \(z\). (b) Express the system of equations you found in part (a) as a matrix equation of the form \(A X=B\). (c) Find the inverse of the coefficient matrix \(A\) and use it to solve the matrix equation in part (b). How much commission does the saleswoman earn on a set of encyclopedias in each grade of binding?

Step-by-Step Solution

Answer

The saleswoman earns $75 for standard, $100 for deluxe, and $250 for leather sets.

Step 1: Setup the system of equations

First, we need to translate the sales data into equations. Let \(x\), \(y\), and \(z\) be the commission for standard, deluxe, and leather sets, respectively. We can create the following equations from the provided data:1. Week One: \(x + y + 2z = 675\)2. Week Two: \(2x + y + z = 600\)3. Week Three: \(x + 2y + z = 625\)These equations represent the total commissions earned based on the sales of different binding grades.

Step 2: Express the system as a matrix equation

We can express the system of linear equations in matrix form \(AX = B\).Let \[A = \begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 1 \ 1 & 2 & 1 \end{bmatrix}\], \[X = \begin{bmatrix} x \ y \ z \end{bmatrix}\], and \[B = \begin{bmatrix} 675 \ 600 \ 625 \end{bmatrix}\].The matrix equation is:\[\begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 1 \ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 675 \ 600 \ 625 \end{bmatrix}\]

Step 3: Find the inverse of the matrix A

To solve for \(X\), we need the inverse of matrix \(A\). The inverse of a matrix \([A]^{-1}\) is found such that \([A]^{-1}A = I\), where \(I\) is the identity matrix. Use a calculator or a formula to compute:\[A^{-1} = \begin{bmatrix} -1 & 1 & 1 \ 1 & 0 & -1 \ 0.5 & -0.5 & 0 \end{bmatrix}\]

Step 4: Solve for X using the inverse matrix

Now, solve the equation \(X = A^{-1}B\) by multiplying \(A^{-1}\) with \(B\):\[X = \begin{bmatrix} -1 & 1 & 1 \ 1 & 0 & -1 \ 0.5 & -0.5 & 0 \end{bmatrix} \begin{bmatrix} 675 \ 600 \ 625 \end{bmatrix} = \begin{bmatrix} 75 \ 100 \ 250 \end{bmatrix}\]This means \(x = 75\), \(y = 100\), and \(z = 250\).

Step 5: Interpret the results

The solution \(x = 75\), \(y = 100\), and \(z = 250\) indicates the commission that the saleswoman earns for each set sold. She earns \\(75 from each standard set, \\)100 from each deluxe set, and \$250 from each leather set.

Key Concepts

Matrix Equation

A matrix equation is a way of expressing a system of linear equations in a compact, simplified manner. It's especially useful when dealing with multiple equations, like in our encyclopedia saleswoman problem. Instead of writing equations separately, we group together the coefficients, variables, and results in matrices.

In our exercise, we have a system of three linear equations:

• Week One: \(x + y + 2z = 675\)
• Week Two: \(2x + y + z = 600\)
• Week Three: \(x + 2y + z = 625\)

These equations can be represented by a single matrix equation, \(AX = B\). Here,

• \(A\) is a coefficient matrix: \[A = \begin{bmatrix} 1 & 1 & 2 \ 2 & 1 & 1 \ 1 & 2 & 1 \end{bmatrix}\]
• \(X\) is a variable matrix: \[X = \begin{bmatrix} x \ y \ z \end{bmatrix}\]
• \(B\) is a result matrix: \[B = \begin{bmatrix} 675 \ 600 \ 625 \end{bmatrix}\]

This equation combines all the given information into a structured format, making it easier to solve using matrix operations.

Inverse Matrix

The inverse of a matrix is the key to unlocking solutions for systems expressed as matrix equations. In our context, the inverse matrix, denoted as \(A^{-1}\), helps us to solve for the matrix \(X\). The essence lies in the property that when \(A^{-1}\) is multiplied by \(A\), it results in an identity matrix \(I\), which when multiplied to any matrix leaves it unchanged. This property is used to solve our equation \(AX = B\).

To find the inverse \(A^{-1}\) of matrix \(A\), we can employ various methods like the adjugate method, row operations, or simply use a calculator for intricate matrices. In our problem, the inverse matrix is:\[A^{-1} = \begin{bmatrix} -1 & 1 & 1 \ 1 & 0 & -1 \ 0.5 & -0.5 & 0 \end{bmatrix}\]To find the values of \(X\), we use the equation \(X = A^{-1}B\). The inverse matrix \(A^{-1}\) facilitates reversing the matrix equation, providing the specific solutions for \(x\), \(y\), and \(z\), which represent the commission earnings for the saleswoman.

Linear Algebra

Linear Algebra is the branch of mathematics focused on vector spaces and linear mappings between these spaces. It plays a fundamental role in solving systems of linear equations, just like our encyclopedia saleswoman's commissions problem.

Using the principles of linear algebra, specifically composition of matrices and operations like finding inverses, allows for efficient solutions in higher dimensions. The concepts of matrix equations and inverse matrices originate from linear algebra. They simplify complex calculations, enabling us to handle multiple equations simultaneously.

The scenarios we solve using systems of equations often model real-world problems. By expressing these in matrix format, linear algebra provides a streamlined method to decode these problems. This approach not only teaches the practical application of math but also illustrates the power of linear transformations and their solutions in real applications like sales commissions and beyond.