Problem 71 - Algebra and Trigonometry

Question

Write each system in the form \(A X=B .\) Then solve the system by entering \(A\) and \(B\) into your graphing utility and computing \(A^{-1} B\) $$ \left\\{\begin{aligned} x-y+z &=-6 \\ 4 x+2 y+z &=9 \\ 4 x-2 y+z &=-3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution for the system of equations is \(x = 1, y = 2, and z = -3\).

Step 1: Formulate Matrix A and B

The first step is to rewrite the system of equations in the matrix form \(AX = B\). For Matrix A, the coefficients of the variables x, y and z in each equation form the rows of the matrix. And Matrix B is formed by the constant term on the right side of each equation. Here, Matrix A and B will be:\(A=\left[\begin{array}{lll}1 & -1 & 1 \ 4 & 2 & 1 \ 4 & -2 & 1\end{array}\right]\), \(B=\left[\begin{array}{c}-6 \ 9 \ -3\end{array}\right]\)

Step 2: Calculate the inverse of Matrix A

Matrix inverse is such that when it is multiplied by original matrix, it results in the identity matrix. We can use calculating function in the graphic utility to get \(A^{-1} = \left[\begin{array}{lll}0 & 0.5 & 0.5 \ -0.333 & 0.167 & -0.167 \ 0.333 & 0.833 & -0.833\end{array}\right]\)

Step 3: Find the Solution of the System

The solution \(X\) of the given system is obtained by multiplying the inverse of Matrix A with Matrix B. Hence, \(X = A^{-1}B\). On computing we get \(X = \left[\begin{array}{c}1 \ 2 \ -3\end{array}\right]\)

Key Concepts

Systems of Linear Equations

A system of linear equations consists of multiple equations with multiple variables that are interdependent. Each equation represents a line, and the solution is the point at which these lines intersect.
For the given system:

• Equation 1: \( x - y + z = -6 \)
• Equation 2: \( 4x + 2y + z = 9 \)
• Equation 3: \( 4x - 2y + z = -3 \)

The goal is to find the values of \(x\), \(y\), and \(z\) that satisfy all equations simultaneously. In matrix algebra, these equations are combined into matrices which help visualize and solve the system more systematically.

Matrix Inversion

Matrix inversion is a crucial technique in solving systems of linear equations. Given a square matrix \(A\), its inverse \(A^{-1}\) is a matrix such that when multiplied with \(A\), the result is the identity matrix, denoted by \(I\).
For example, considering the matrix \(A\):\[A=\left[\begin{array}{ccc}1 & -1 & 1 \4 & 2 & 1 \4 & -2 & 1\end{array}\right]\]
Its inverse \(A^{-1}\) is calculated:\[A^{-1} = \left[\begin{array}{ccc}0 & 0.5 & 0.5 \-0.333 & 0.167 & -0.167 \0.333 & 0.833 & -0.833\end{array}\right]\]
This inversion process often involves advanced computational techniques and is facilitated using graphing utilities or algebraic software due to its complexity.

Graphing Utility Solutions

Using a graphing utility can greatly simplify the process of solving complex systems of equations and calculating matrix inverses.
A graphing utility allows input of matrices directly and utilizes its built-in functions to compute necessary operations like matrix inversion. For our system, once inputs are made as described before:

• Matrix \(A\) is input with its coefficients
• Matrix \(B\) is input with its constants
• The utility computes \(A^{-1}\) and then multiplies it with \(B\)

This results in the solution matrix \(X\), where each entry corresponds to one of the variables in the system. In this case, matrix \(X\) was found to be:\[X = \left[\begin{array}{c}1 \2 \-3\end{array}\right]\]Thus, \(x = 1\), \(y = 2\), and \(z = -3\) are the solutions to the system.