Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Title: Solving Systems of Linear Equations Using the Inverse Matrix Method**
**Objective:**
Learn how to solve a system of linear equations by applying the inverse matrix method.
**Given System of Equations:**
1. \( x + y - 3z = 2 \)
2. \( 2x + 3y + z = 0 \)
3. \( 3x + 4y - z = 1 \)
**Methodology:**
To solve this system, we will use matrices. The system can be written in matrix form as \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 1 & -3 \\ 2 & 3 & 1 \\ 3 & 4 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \]
The solution involves finding the inverse of matrix A, denoted \( A^{-1} \), such that:
\[ X = A^{-1}B \]
**Conclusion:**
By calculating \( A^{-1} \) and multiplying it by matrix \( B \), you will obtain the values for the variables \( x \), \( y \), and \( z \) that satisfy all three equations.
**Note:**
This technique is applicable when the matrix \( A \) is invertible, meaning its determinant is non-zero. If \( A \) is not invertible, other methods such as Gaussian elimination or matrix decomposition may be used to solve the system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb497b8c4-4d07-4c4c-ac55-469da516b240%2Fbc6de926-02dd-4f95-8e17-0517c327ca2c%2Fztr0cws_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving Systems of Linear Equations Using the Inverse Matrix Method**
**Objective:**
Learn how to solve a system of linear equations by applying the inverse matrix method.
**Given System of Equations:**
1. \( x + y - 3z = 2 \)
2. \( 2x + 3y + z = 0 \)
3. \( 3x + 4y - z = 1 \)
**Methodology:**
To solve this system, we will use matrices. The system can be written in matrix form as \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 1 & -3 \\ 2 & 3 & 1 \\ 3 & 4 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \]
The solution involves finding the inverse of matrix A, denoted \( A^{-1} \), such that:
\[ X = A^{-1}B \]
**Conclusion:**
By calculating \( A^{-1} \) and multiplying it by matrix \( B \), you will obtain the values for the variables \( x \), \( y \), and \( z \) that satisfy all three equations.
**Note:**
This technique is applicable when the matrix \( A \) is invertible, meaning its determinant is non-zero. If \( A \) is not invertible, other methods such as Gaussian elimination or matrix decomposition may be used to solve the system.
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