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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![### Solving Systems of Equations Using Gaussian Elimination
---
**Problem Statement:**
Solve the system of equations using Gaussian Elimination:
1. \( x - 2y + 3z = 1 \)
2. \( x + 2y - z = 13 \)
3. \( 3x + 2y - 5z = 3 \)
**Hint:** Find the Row Echelon Form
---
**Solution Steps:**
1. **Write the system as a matrix:**
\[
\begin{bmatrix}
1 & -2 & 3 & | & 1 \\
1 & 2 & -1 & | & 13 \\
3 & 2 & -5 & | & 3
\end{bmatrix}
\]
2. **Perform Gaussian Elimination to reach Row Echelon Form:**
- Subtract the first row from the second row:
\[
(-x + 2y - z = 13) - (x - 2y + 3z = 1) \rightarrow 4y - 4z = 12
\]
Simplifying gives:
\[
4y - 4z = 12 \quad \Rightarrow \quad y - z = 3
\]
- Subtract three times the first row from the third row:
\[
(3x + 2y - 5z = 3) - 3(x - 2y + 3z = 1) \rightarrow -4y + 14z = 0
\]
Simplifying gives:
\[
-4y + 14z = 0 \quad \Rightarrow \quad 2y - 7z = 0 \quad
\]
3. **Further simplification of the equations results in:**
\[
x + y + z = 1
\]
\[
y - z = 2
\]
\[
z = 3
\]
4. **Final Solution:**
\[
\begin{cases}
x + y + z = 1 \\
y - z = 2 \\
z = 3
\end{cases}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64a9bba9-6d54-4904-907b-a62783e46f3d%2Fed090c41-4bf5-4ef2-a00a-5502b593aeb5%2F5u9hetg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Systems of Equations Using Gaussian Elimination
---
**Problem Statement:**
Solve the system of equations using Gaussian Elimination:
1. \( x - 2y + 3z = 1 \)
2. \( x + 2y - z = 13 \)
3. \( 3x + 2y - 5z = 3 \)
**Hint:** Find the Row Echelon Form
---
**Solution Steps:**
1. **Write the system as a matrix:**
\[
\begin{bmatrix}
1 & -2 & 3 & | & 1 \\
1 & 2 & -1 & | & 13 \\
3 & 2 & -5 & | & 3
\end{bmatrix}
\]
2. **Perform Gaussian Elimination to reach Row Echelon Form:**
- Subtract the first row from the second row:
\[
(-x + 2y - z = 13) - (x - 2y + 3z = 1) \rightarrow 4y - 4z = 12
\]
Simplifying gives:
\[
4y - 4z = 12 \quad \Rightarrow \quad y - z = 3
\]
- Subtract three times the first row from the third row:
\[
(3x + 2y - 5z = 3) - 3(x - 2y + 3z = 1) \rightarrow -4y + 14z = 0
\]
Simplifying gives:
\[
-4y + 14z = 0 \quad \Rightarrow \quad 2y - 7z = 0 \quad
\]
3. **Further simplification of the equations results in:**
\[
x + y + z = 1
\]
\[
y - z = 2
\]
\[
z = 3
\]
4. **Final Solution:**
\[
\begin{cases}
x + y + z = 1 \\
y - z = 2 \\
z = 3
\end{cases}
\]
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