use the reduced row echelon form above to solve the system 2х + 2у — z —х — 2у + 2z -3 x + y – z 2
use the reduced row echelon form above to solve the system 2х + 2у — z —х — 2у + 2z -3 x + y – z 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![In this problem, you are given a system of linear equations represented by the following augmented matrix, which is transformed into its reduced row echelon form (RREF):
Initial Matrix:
\[
\begin{bmatrix}
2 & 2 & -1 & 0 \\
-1 & -2 & 2 & -3 \\
1 & 1 & -1 & 2
\end{bmatrix}
\quad \sim \quad
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -3 \\
0 & 0 & 1 & -4
\end{bmatrix}
\]
The reduced row echelon form allows us to easily read the solution to the following system of equations:
\[
\begin{cases}
2x + 2y - z = 0 \\
-x - 2y + 2z = -3 \\
x + y - z = 2
\end{cases}
\]
Using the RREF, we deduce the solutions to the variables \( x \), \( y \), and \( z \):
\[
x = 1, \quad y = -3, \quad z = -4
\]
If necessary, parameterize your answer using the free variables of the system.
Enter your answers in the matrix below:
\[
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
1 \\
-3 \\
-4
\end{bmatrix}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F31330aa5-a64b-4d2f-80e6-6ea74abb2a34%2Fad6ecfab-ffa1-4508-9b6d-04aecd647775%2Fmjfllet_processed.png&w=3840&q=75)
Transcribed Image Text:In this problem, you are given a system of linear equations represented by the following augmented matrix, which is transformed into its reduced row echelon form (RREF):
Initial Matrix:
\[
\begin{bmatrix}
2 & 2 & -1 & 0 \\
-1 & -2 & 2 & -3 \\
1 & 1 & -1 & 2
\end{bmatrix}
\quad \sim \quad
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -3 \\
0 & 0 & 1 & -4
\end{bmatrix}
\]
The reduced row echelon form allows us to easily read the solution to the following system of equations:
\[
\begin{cases}
2x + 2y - z = 0 \\
-x - 2y + 2z = -3 \\
x + y - z = 2
\end{cases}
\]
Using the RREF, we deduce the solutions to the variables \( x \), \( y \), and \( z \):
\[
x = 1, \quad y = -3, \quad z = -4
\]
If necessary, parameterize your answer using the free variables of the system.
Enter your answers in the matrix below:
\[
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
1 \\
-3 \\
-4
\end{bmatrix}
\]
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