Consider the following system of linear equations. 4x-2y -z = 14 +2z = 26 =-3 5x 2x +y Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
Consider the following system of linear equations. 4x-2y -z = 14 +2z = 26 =-3 5x 2x +y Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Title: Solving a System of Linear Equations Using Row-Echelon Form**
**Introduction:**
Consider the following system of linear equations:
\[ 4x - 2y - z = 14 \]
\[ 5x + 2z = 26 \]
\[ 2x + y = -3 \]
Our goal is to solve this system by transforming the equations into a reduced row-echelon form using matrix operations. We denote the first, second, and third rows as \( R_1 \), \( R_2 \), and \( R_3 \) respectively. The arrow notation (\(\rightarrow\)) indicates that the expression on the left replaces the expression on the right.
---
**Step-by-Step Solution:**
1. **Initial Matrix Setup:**
The system is represented by the augmented matrix:
\[
\begin{bmatrix}
4 & -2 & -1 & | & 14 \\
5 & 0 & 2 & | & 26 \\
2 & 1 & 0 & | & -3 \\
\end{bmatrix}
\]
2. **Row Operations:**
**(1)** Scale \( R_1 \):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & | & \frac{7}{2} \\
5 & 0 & 2 & | & 26 \\
2 & 1 & 0 & | & -3 \\
\end{bmatrix}
\]
**(2)** Update \( R_2 \) and \( R_3 \):
\[
(-5) \cdot R_1 + R_2 \rightarrow R_2
\]
\[
R_1 + R_3 \rightarrow R_3
\]
Resulting matrix:
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & | & \frac{7}{2} \\
0 & 1 & \frac{13}{4} & | & \frac{17}{2} \\
0 & 2 & 1 & | & -10 \\
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2039606c-ddf8-4dfc-99c4-5a13e28c47b4%2F99e37430-0862-4ec6-a317-dd20be914a0c%2Fsrmjdd7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving a System of Linear Equations Using Row-Echelon Form**
**Introduction:**
Consider the following system of linear equations:
\[ 4x - 2y - z = 14 \]
\[ 5x + 2z = 26 \]
\[ 2x + y = -3 \]
Our goal is to solve this system by transforming the equations into a reduced row-echelon form using matrix operations. We denote the first, second, and third rows as \( R_1 \), \( R_2 \), and \( R_3 \) respectively. The arrow notation (\(\rightarrow\)) indicates that the expression on the left replaces the expression on the right.
---
**Step-by-Step Solution:**
1. **Initial Matrix Setup:**
The system is represented by the augmented matrix:
\[
\begin{bmatrix}
4 & -2 & -1 & | & 14 \\
5 & 0 & 2 & | & 26 \\
2 & 1 & 0 & | & -3 \\
\end{bmatrix}
\]
2. **Row Operations:**
**(1)** Scale \( R_1 \):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & | & \frac{7}{2} \\
5 & 0 & 2 & | & 26 \\
2 & 1 & 0 & | & -3 \\
\end{bmatrix}
\]
**(2)** Update \( R_2 \) and \( R_3 \):
\[
(-5) \cdot R_1 + R_2 \rightarrow R_2
\]
\[
R_1 + R_3 \rightarrow R_3
\]
Resulting matrix:
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & | & \frac{7}{2} \\
0 & 1 & \frac{13}{4} & | & \frac{17}{2} \\
0 & 2 & 1 & | & -10 \\
\
![Certainly! Below is a transcription of the image content for an educational website, including details of the matrix transformations and blanks to be filled in.
---
**Matrix Row Operations and Solutions**
**(2)**
1. Perform the operation \((-5) \cdot R_1 + R_2 \to R_2\):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & \vrule & \frac{7}{2} \\
0 & \frac{5}{2} & \frac{13}{4} & \vrule & \frac{17}{2} \\
0 & 2 & \frac{1}{2} & \vrule & -10
\end{bmatrix}
\]
2. Perform the operation \(\boxed{\phantom{a}} \cdot R_1 + R_3 \to R_3\).
**(3)**
1. Transform \(R_2\) using: \(\boxed{\phantom{a}} \cdot R_2 \to R_2\):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & \vrule & \frac{7}{2} \\
0 & 1 & \frac{13}{10} & \vrule & \frac{17}{5} \\
0 & 2 & \frac{1}{2} & \vrule & -10
\end{bmatrix}
\]
**(4)**
1. Perform the operation \(\left(\frac{1}{2}\right) \cdot R_2 + R_1 \to R_1\):
\[
\begin{bmatrix}
1 & 0 & \frac{2}{5} & \vrule & \frac{26}{5} \\
0 & 1 & \frac{13}{10} & \vrule & \frac{17}{5} \\
0 & 0 & -\frac{21}{10} & \vrule & -\frac{84}{5}
\end{bmatrix}
\]
2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2039606c-ddf8-4dfc-99c4-5a13e28c47b4%2F99e37430-0862-4ec6-a317-dd20be914a0c%2F20kbahn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Below is a transcription of the image content for an educational website, including details of the matrix transformations and blanks to be filled in.
---
**Matrix Row Operations and Solutions**
**(2)**
1. Perform the operation \((-5) \cdot R_1 + R_2 \to R_2\):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & \vrule & \frac{7}{2} \\
0 & \frac{5}{2} & \frac{13}{4} & \vrule & \frac{17}{2} \\
0 & 2 & \frac{1}{2} & \vrule & -10
\end{bmatrix}
\]
2. Perform the operation \(\boxed{\phantom{a}} \cdot R_1 + R_3 \to R_3\).
**(3)**
1. Transform \(R_2\) using: \(\boxed{\phantom{a}} \cdot R_2 \to R_2\):
\[
\begin{bmatrix}
1 & -\frac{1}{2} & -\frac{1}{4} & \vrule & \frac{7}{2} \\
0 & 1 & \frac{13}{10} & \vrule & \frac{17}{5} \\
0 & 2 & \frac{1}{2} & \vrule & -10
\end{bmatrix}
\]
**(4)**
1. Perform the operation \(\left(\frac{1}{2}\right) \cdot R_2 + R_1 \to R_1\):
\[
\begin{bmatrix}
1 & 0 & \frac{2}{5} & \vrule & \frac{26}{5} \\
0 & 1 & \frac{13}{10} & \vrule & \frac{17}{5} \\
0 & 0 & -\frac{21}{10} & \vrule & -\frac{84}{5}
\end{bmatrix}
\]
2.
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