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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![The image displays a matrix in reduced row echelon form and a question. Here is the detailed transcription and explanation:
The matrix given is:
\[
\begin{bmatrix}
1 & 0 & -3 & \vline & 6 \\
0 & 1 & 5 & \vline & 9 \\
0 & 0 & 0 & \vline & 0 \\
\end{bmatrix}
\]
This is an augmented matrix representing a system of linear equations. The vertical line separates the coefficients of the variables from the constants on the right side of the equations.
The system is:
1. \( x_1 - 3x_3 = 6 \)
2. \( x_2 + 5x_3 = 9 \)
The third row indicates no additional equation because it is all zeros.
The question accompanying the matrix is:
"What is the final solution? _______________________________"
This implies the need for solving the system for \( x_1 \), \( x_2 \), and \( x_3 \).
To solve this, express \( x_1 \) and \( x_2 \) in terms of \( x_3 \), which can be any real number due to row of zeros:
- \( x_1 = 6 + 3x_3 \)
- \( x_2 = 9 - 5x_3 \)
Thus, the solution set can be expressed in terms of \( x_3 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4a6764b-ec66-4cf8-9984-52f9ddecdfb1%2F468ac4d8-6d56-41c0-8c68-81564273e83c%2Fv61wb1c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image displays a matrix in reduced row echelon form and a question. Here is the detailed transcription and explanation:
The matrix given is:
\[
\begin{bmatrix}
1 & 0 & -3 & \vline & 6 \\
0 & 1 & 5 & \vline & 9 \\
0 & 0 & 0 & \vline & 0 \\
\end{bmatrix}
\]
This is an augmented matrix representing a system of linear equations. The vertical line separates the coefficients of the variables from the constants on the right side of the equations.
The system is:
1. \( x_1 - 3x_3 = 6 \)
2. \( x_2 + 5x_3 = 9 \)
The third row indicates no additional equation because it is all zeros.
The question accompanying the matrix is:
"What is the final solution? _______________________________"
This implies the need for solving the system for \( x_1 \), \( x_2 \), and \( x_3 \).
To solve this, express \( x_1 \) and \( x_2 \) in terms of \( x_3 \), which can be any real number due to row of zeros:
- \( x_1 = 6 + 3x_3 \)
- \( x_2 = 9 - 5x_3 \)
Thus, the solution set can be expressed in terms of \( x_3 \).
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