Use row reduction to solve the following system. = 3 X1 - 4x₂ + 2x3 3x2 + 5x3 = -7 -2x₁ + 8x₂ - 4x3 = -3
Use row reduction to solve the following system. = 3 X1 - 4x₂ + 2x3 3x2 + 5x3 = -7 -2x₁ + 8x₂ - 4x3 = -3
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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![**Solving System of Linear Equations using Row Reduction**
Use row reduction to solve the following system:
\[
\begin{cases}
x_1 - 4x_2 + 2x_3 = 3 \\
3x_2 + 5x_3 = -7 \\
-2x_1 + 8x_2 - 4x_3 = -3
\end{cases}
\]
In order to solve this system using row reduction, we follow these steps:
1. **Write the augmented matrix** for the system of equations.
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
-2 & 8 & -4 & -3
\end{array}\right]
\]
2. **Apply the row operations** to reduce the matrix to row echelon form and then to reduced row echelon form.
### Step-by-Step Process:
- Start with the initial augmented matrix:
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
-2 & 8 & -4 & -3
\end{array}\right]
\]
- First, we can eliminate the \(x_1\) terms from the third row:
- Multiply the first row by 2 and add to the third row:
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
0 & 0 & 0 & 3
\end{array}\right]
\]
- Next, we continue row reduction to simplify the second row, ensuring the leading coefficient in each row is 1 where possible.
Finally, interpret the resulting row-echelon form to solve for \(x_1\), \(x_2\), and \(x_3\).
In depth, the actual row reduction steps involve further elementary row operations (like swapping, multiplying, adding, and subtracting rows) to convert the matrix into reduced row echelon form from which the solutions can be read directly.
*Note:* The above steps outline the general process. For full row reduction, each step would include further individual operations until the matrix](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc7f96b0-6dc0-4972-bfe0-9c219792bf84%2F1530e905-6de1-4b4e-9957-71c4cb37c679%2F0gls2v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving System of Linear Equations using Row Reduction**
Use row reduction to solve the following system:
\[
\begin{cases}
x_1 - 4x_2 + 2x_3 = 3 \\
3x_2 + 5x_3 = -7 \\
-2x_1 + 8x_2 - 4x_3 = -3
\end{cases}
\]
In order to solve this system using row reduction, we follow these steps:
1. **Write the augmented matrix** for the system of equations.
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
-2 & 8 & -4 & -3
\end{array}\right]
\]
2. **Apply the row operations** to reduce the matrix to row echelon form and then to reduced row echelon form.
### Step-by-Step Process:
- Start with the initial augmented matrix:
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
-2 & 8 & -4 & -3
\end{array}\right]
\]
- First, we can eliminate the \(x_1\) terms from the third row:
- Multiply the first row by 2 and add to the third row:
\[
\left[\begin{array}{ccc|c}
1 & -4 & 2 & 3 \\
0 & 3 & 5 & -7 \\
0 & 0 & 0 & 3
\end{array}\right]
\]
- Next, we continue row reduction to simplify the second row, ensuring the leading coefficient in each row is 1 where possible.
Finally, interpret the resulting row-echelon form to solve for \(x_1\), \(x_2\), and \(x_3\).
In depth, the actual row reduction steps involve further elementary row operations (like swapping, multiplying, adding, and subtracting rows) to convert the matrix into reduced row echelon form from which the solutions can be read directly.
*Note:* The above steps outline the general process. For full row reduction, each step would include further individual operations until the matrix
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