Problem 50 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} (\log 2) x+(\ln 3) y+(\ln 4) z &=1 \\ (\ln 3) x+(\log 2) y+(\ln 8) z &=5 \\ (\log 12) x+(\ln 4) y+(\ln 8) z &=9 \end{aligned}$$

Step-by-Step Solution

Answer

Calculate \( A^{-1} \) and multiply by \( B \) to find \( x, y, z \).

Step 1: Write the System in Matrix Form

We start by expressing the given system of equations in matrix form. Let the coefficient matrix be \( A \), the variable matrix be \( X \), and the constant matrix be \( B \).The system is:\[ \begin{bmatrix} \log 2 & \ln 3 & \ln 4 \ \ln 3 & \log 2 & \ln 8 \ \log 12 & \ln 4 & \ln 8 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 5 \ 9 \end{bmatrix} \]Here, \( A = \begin{bmatrix} \log 2 & \ln 3 & \ln 4 \ \ln 3 & \log 2 & \ln 8 \ \log 12 & \ln 4 & \ln 8 \end{bmatrix} \), \( X = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( B = \begin{bmatrix} 1 \ 5 \ 9 \end{bmatrix} \).

Step 2: Calculate the Inverse of Matrix A

The inverse of a matrix \( A \) (denoted as \( A^{-1} \)) can be found as long as \( A \) is a square matrix and its determinant is non-zero. Calculate \( A^{-1} \) using: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]Use matrix computation or software to compute this, as manual calculation is beyond this scope here.

Step 3: Solve for Matrix X

Once the inverse is computed, the matrix equation \( A \cdot X = B \) can be solved as:\[ X = A^{-1} \cdot B \]Multiply the inverse of \( A \) with matrix \( B \) to find \( X \):\( X = \begin{bmatrix} x \ y \ z \end{bmatrix} \).

Step 4: Interpret the Solution

After computing \( X = A^{-1} \cdot B \), we arrive at numerical values for \( x \), \( y \), and \( z \). These represent the solution to the system, i.e., the values of \( x \), \( y \), and \( z \) that satisfy all three equations simultaneously.

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In this problem, the unknowns are the variables \( x \), \( y \), and \( z \). Each equation in the system represents a constraint, and the goal is to find the values of these unknowns that satisfy all equations simultaneously.
For example, in the given system, you have:

• \((\log 2) x + (\ln 3) y + (\ln 4) z =1 \)
• \((\ln 3)x + (\log 2) y + (\ln 8) z = 5 \)
• \((\log 12) x + (\ln 4) y + (\ln 8) z = 9 \)

The objective is to find values for \( x \), \( y \), and \( z \) that make these equations true at the same time. Solving systems of equations can be approached in various ways, including graphically, algebraically, or using matrix methods as highlighted in this exercise.

Matrix Algebra

Matrix algebra is a branch of mathematics dealing with matrices, or rectangular grids of numbers, which can represent systems of linear equations. In this context, you'll often see matrices involved in solving systems of equations, which is done using operations like matrix addition, multiplication, and finding inverses.
The matrix form of a system of equations condenses the problem, making it easier to visualize and manipulate. Here's how the given system appears in matrix form:\[\begin{bmatrix} \log 2 & \ln 3 & \ln 4 \ \ln 3 & \log 2 & \ln 8 \ \log 12 & \ln 4 & \ln 8 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 5 \ 9 \end{bmatrix}\]This matrix equation represents the same information as the original set of equations but allows you to use powerful matrix operations, like finding inverses, to simplify solving.

Logarithms

Logarithms are the inverse operations to exponentiation. They help undo what exponentiation does. If you have a number expressed in terms of a power, a logarithm lets you find that power. For example, \( \log_{10} (100) = 2 \) because 10 squared is 100.
In the context of this exercise, logarithms of different bases appear in the coefficients of the equations. The \( \log \) refers to base 10, and \( \ln \) refers to the natural logarithm, base \( e \). Knowing how to manipulate and understand these is crucial since they are part of the matrix coefficient values.

• \( \log 2 \) indicates a logarithm base 10 of 2.
• \( \ln 3 \) suggests a natural logarithm (base \( e \)) of 3.

It's essential to be comfortable converting and using logarithms to make sense of their roles in the equations and their manipulations.

Linear Algebra

Linear algebra is the field of mathematics that studies vectors, matrices, and linear transformations. It serves as a foundation for handling systems of linear equations like the one in this exercise. One of the most vital tools in linear algebra is the method of finding the inverse of a matrix.
This method is useful for solving equations presented in matrix form. If you have a system \( A \times X = B \), you can find \( X \) by calculating \( A^{-1} \) and then multiplying it by \( B \) as per:\[ X = A^{-1} \cdot B \]Finding \( A^{-1} \) helps isolate the variable matrix \( X \). It's only possible if the determinant of \( A \) is not zero, which ensures that the inverse exists. Thus, linear algebra offers systematic tools and strategies to deal with complex systems of equations using matrices.