Problem 49 - Algebra and Trigonometry

Question

Sales Commissions An encyclopedia saleswoman works for a company that offers three different grades of bindings for its encyclopedias: standard, deluxe, and leather. For each set that 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 $200 for leather sets.

Step 1: Define variables and set up equations

Assign variables: let \(x\) be the commission for a standard set, \(y\) for a deluxe set, and \(z\) for a leather set. Based on the sales information:- First week: \(x + y + 2z = 675\)- Second week: \(2x + y + z = 600\)- Third week: \(x + 2y + z = 625\)

Step 2: Write the system of equations as A X = B

The coefficients of \(x\), \(y\), and \(z\) in the three equations can be represented in matrix form:\[A = \begin{bmatrix}1 & 1 & 2 \2 & 1 & 1 \1 & 2 & 1\end{bmatrix}, \ X = \begin{bmatrix}x \y \z\end{bmatrix}, \ B = \begin{bmatrix}675 \600 \625\end{bmatrix}\]Thus, we have the matrix equation \(AX = B\).

Step 3: Find inverse of matrix A

To solve the matrix equation using the inverse, first find the inverse of matrix \(A\), \(A^{-1}\):The determinant of \(A\) is \[\text{det}(A) = 1(1-2) - 1(2-1) + 2(4-1) = -1 - 1 + 6 = 4\]Calculate \(A^{-1}\) as follows:\[A^{-1} = \frac{1}{4} \begin{bmatrix}1 & -1 & 2 \-2 & 1 & 0 \1 & 0 & -1\end{bmatrix} = \begin{bmatrix}0.25 & -0.25 & 0.5 \-0.5 & 0.25 & 0 \0.25 & 0 & -0.25\end{bmatrix}\]

Step 4: Solve the matrix equation using A^{-1}

Substitute \(A^{-1}\) into the equation to find \(X = A^{-1}B\):\[X = \begin{bmatrix}0.25 & -0.25 & 0.5 \-0.5 & 0.25 & 0 \0.25 & 0 & -0.25\end{bmatrix} \begin{bmatrix}675 \600 \625\end{bmatrix}\]Calculate \(X\):\[X = \begin{bmatrix}0.25(675) - 0.25(600) + 0.5(625) \-0.5(675) + 0.25(600) + 0 \0.25(675) + 0 - 0.25(625)\end{bmatrix}\]\[X = \begin{bmatrix}75 \100 \200\end{bmatrix}\]

Step 5: Interpret the solution

The solution \(\begin{bmatrix} 75 \ 100 \ 200 \end{bmatrix}\) means that the saleswoman earns \(\\(75\) commission for a standard set, \(\\)100\) for a deluxe set, and \(\$200\) for a leather set.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In this exercise, we have three equations and three variables: \(x\), \(y\), and \(z\) which represent the commission earned on standard, deluxe, and leather sets respectively. The equations provided are derived from the commission information over three weeks:
1. \(x + y + 2z = 675\) - First week's commission.
2. \(2x + y + z = 600\) - Second week's commission.
3. \(x + 2y + z = 625\) - Third week's commission.

Solving this system of equations allows us to find the individual commissions for each type of binding. The key to solving systems of equations is either substitution, elimination, or matrix representation, like in this example.

• This problem illustrates how real-life problems involving sold products and commissions can be expressed as mathematical equations.
• Understanding the system of equations is crucial for representing such data and deriving solutions using algebraic methods.

Matrix Inversion

Matrix inversion is a technique from algebra that involves finding another matrix which, when multiplied by the original matrix, yields the identity matrix. In our case, the matrix to invert is \(A\), the coefficient matrix of the linear system. Not all matrices can be inverted, but for those that can, it is a powerful method for solving systems of linear equations.

The inverse of matrix \(A\), denoted as \(A^{-1}\), allows us to solve the system expressed in the matrix equation form \(AX = B\). By calculating \(A^{-1}\), we can find \(X\) through the equation \(X = A^{-1}B\).

• The determinant of \(A\) must be non-zero for \(A\) to have an inverse.
• Finding the inverse involves creating the matrix of minors, cofactors, adjugate, and then dividing by the determinant.

Matrix inversion offers a systematic approach to solving linear equations, though simpler methods might be more efficient for smaller systems.

Linear Algebra

Linear algebra is the branch of mathematics that studies vectors, matrices, and linear transformations. It is fundamental to systems of linear equations like the one in our problem. Linear algebra provides multiple strategies to solve these systems, whether by using determinants and inverses or leveraging computational algorithms.

In this problem, linear algebra helps in converting the sales commission problem into a matrix form, allowing us to engage various techniques to solve the equations. Understanding matrices and their operations, including multiplication and inversion, is central to solving linear systems efficiently.

• Linear algebra simplifies solving complex systems that would otherwise be difficult to solve manually.
• Applications extend beyond homework problems to include computer graphics, engineering, physics, and more.

While the topic can seem abstract, it is incredibly practical, providing tools to model and solve real-world problems.

Matrix Representation

Matrix representation is a way of expressing a system of linear equations in a compact form using matrices. In this exercise, we describe the sales commission system as a matrix equation \(AX = B\). Here:
\[A = \begin{bmatrix}1 & 1 & 2 \ 2 & 1 & 1 \ 1 & 2 & 1\end{bmatrix}, \quad X = \begin{bmatrix}x \ y \ z\end{bmatrix}, \quad B = \begin{bmatrix}675 \ 600 \ 625\end{bmatrix}\]

This form allows mathematicians and students to apply various matrix operations and integral properties for solutions. Matrix representation not only makes the handling of multiple equations cleaner but also facilitates the use of matrix-specific solutions like Gaussian elimination and Cramer's Rule when applicable.

• Transforming problems into matrices can make calculations more organized and systematic.
• It is a stepping stone to advanced math topics like eigenvalues, eigenvectors, and linear transformations.

For students, mastering matrix representation is an exciting leap towards modern applications in technology and science.