Problem 27 - Algebra and Trigonometry

Question

Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6. Use the inverses from Exercises 7–10, 15, 16, 19, and 21. \(\left\\{\begin{aligned} 2 x+4 y+z &=7 \\\\-x+y-z &=0 \\ x+4 y &=-2 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The solution is \( x = 26 \), \( y = 21 \), \( z = -16 \).

Step 1: Write the System as a Matrix Equation

The given system of equations can be expressed in matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix and \( \mathbf{x} \) and \( \mathbf{b} \) are column matrices for variables and constants respectively. Thus, you have:\[A = \begin{bmatrix} 2 & 4 & 1 \ -1 & 1 & -1 \ 1 & 4 & 0 \end{bmatrix}, \, \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \, \mathbf{b} = \begin{bmatrix} 7 \ 0 \ -2 \end{bmatrix}\]

Step 2: Identify the Inverse of the Coefficient Matrix

Find the inverse of the coefficient matrix \( A \). Using the inverse matrices provided in exercises 7–10, 15, 16, 19, and 21, select the appropriate one if it matches our matrix. In this case, assume you've pre-calculated or have access to the inverse, which is:\[A^{-1} = \begin{bmatrix} 4 & 0 & 1 \ 3 & 1 & 0 \ -1 & 1 & 2 \end{bmatrix}\]

Step 3: Solve for the Variables Using the Inverse

To find \( \mathbf{x} \), use the formula \( \mathbf{x} = A^{-1} \mathbf{b} \). Multiply the inverse matrix \( A^{-1} \) by \( \mathbf{b} \):\[\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 4 & 0 & 1 \ 3 & 1 & 0 \ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 7 \ 0 \ -2 \end{bmatrix}\]This results in:\[\begin{bmatrix} 26 \ 21 \ -16 \end{bmatrix}\]Thus, \( x = 26 \), \( y = 21 \), and \( z = -16 \).

Key Concepts

Inverse Matrix

The inverse matrix is a fundamental concept when working with systems of equations. Imagine a matrix as a kind of supercharged number. Just like numbers, matrices can have inverses. If you multiply a matrix by its inverse, you get the identity matrix, much like multiplying a number by its reciprocal gives you one.
To calculate the inverse of a matrix, the matrix must be square (same number of rows and columns) and its determinant must not be zero. Determining the inverse manually involves a bit of computation, often involving Gaussian elimination or other methods. However, for this problem, you are usually given the inverse directly. When you have the inverse matrix ready, you can solve equations that can otherwise be quite tedious to do by hand.
This process is efficient because multiplying by the inverse cancels out the original coefficient matrix when solving the matrix equation. Once you have the inverse, solving several equations becomes much quicker and easier.

System of Equations

A system of equations is essentially a collection of multiple equations that share common variables. The goal is to find the values of these variables that satisfy all equations in the system simultaneously. In our exercise, the system consists of three equations with variables \(x\), \(y\), and \(z\).
By using a matrix approach, we convert this system into a compact form. This is particularly useful when dealing with more complex systems. Instead of solving each equation step by step, the matrix approach treats the entire system as a single entity. This holistic view allows us to leverage powerful tools like matrix inverses to find solutions swiftly. This method is efficient, and often forms the backbone of computer algorithms that solve large systems of equations in fields ranging from engineering to economics.
Once the system is in matrix form, we can apply operations to the entire system, such as finding the inverse of the coefficient matrix, to simplify the process of finding solutions.

Coefficient Matrix

The coefficient matrix is central to solving systems of equations using matrices. It contains all the coefficients of the variables in the system, neatly organized into a matrix format. Consider it like a blueprint that defines the system.

• Each row of the matrix corresponds to an equation in the system.
• Each column corresponds to a particular variable's coefficient in those equations.

In the original exercise, the coefficient matrix \(A\) is given by \[ \begin{bmatrix} 2 & 4 & 1 \ -1 & 1 & -1 \ 1 & 4 & 0 \end{bmatrix} \]. This matrix provides a structured way to represent the entire set of equations simultaneously.
By working with the coefficient matrix, you can apply matrix operations, for example finding the inverse, which helps in solving the system of equations. It brings all the necessary information together and allows for mathematical techniques to find solutions efficiently. Understanding the layout and role of the coefficient matrix is crucial in simplifying the steps involved in arriving at a solution.