turn in row echelon form: −2x1	 + 	5x2	 − 	10x3	 = 	−5
x1	 − 	2x2	 + 	3x3	 = 	7
7x1	 − 	17x2	 + 	34x3	 = 	15

To put the given system of equations into **row echelon form**, we will perform Gaussian elimination. Here's the system of equations:

\[
\begin{aligned}
-2x_1 + 5x_2 - 10x_3 &= -5 \\
x_1 - 2x_2 + 3x_3 &= 7 \\
7x_1 - 17x_2 + 34x_3 &= 15
\end{aligned}
\]

This can be written as an augmented matrix:

\[
\begin{pmatrix}
-2 & 5 & -10 & | & -5 \\
1 & -2 & 3 & | & 7 \\
7 & -17 & 34 & | & 15
\end{pmatrix}
\]

### Step 1: Make the leading coefficient of the first row 1
We can achieve this by multiplying the second row (which already has a 1 in the first column) by 2 and adding it to the first row to eliminate the \(-2\).

1. Multiply Row 2 by 2 and add to Row 1:
\[
\text{New Row 1} = (-2, 5, -10, | -5) + (2, -4, 6, | 14) = (0, 1, -4, | 9)
\]

Updated matrix:
\[
\begin{pmatrix}
0 & 1 & -4 & | & 9 \\
1 & -2 & 3 & | & 7 \\
7 & -17 & 34 & | & 15
\end{pmatrix}
\]

### Step 2: Eliminate the first column in Row 3
We want to eliminate the 7 in the first column of Row 3 by subtracting \(7 \times \text{Row 2}\) from Row 3.

\[
\text{New Row 3} = (7, -17, 34, | 15) - 7(1, -2, 3, | 7) = (0, -3, 13, | -34)
\]

Updated matrix:
\[
\begin{pmatrix}
0 & 1 & -4 & | & 9 \\
1 & -2 & 3 & | & 7 \\
0 & -3 & 13 & | & -34
\end{pmatrix}
\]

### Step 3: Eliminate the second column in Row 3
Now, we can eliminate the \(-3\) in the second column of Row 3 by adding \(3 \times \text{Row 1}\) to Row 3.

\[
\text{New Row 3} = (0, -3, 13, | -34) + 3(0, 1, -4, | 9) = (0, 0, 1, | -7)
\]

Updated matrix:
\[
\begin{pmatrix}
0 & 1 & -4 & | & 9 \\
1 & -2 & 3 & | & 7 \\
0 & 0 & 1 & | & -7
\end{pmatrix}
\]

### Step 4: Simplify Row 1 and Row 2
- Row 1 can be simplified by adding \(4 \times \text{Row 3}\) to eliminate the \(-4\) in the third column of Row 1:
\[
\text{New Row 1} = (0, 1, -4, | 9) + 4(0, 0, 1, | -7) = (0, 1, 0, | -19)
\]
- Row 2 can be simplified by subtracting \(3 \times \text{Row 3}\) from Row 2 to eliminate the 3 in the third column of Row 2:
\[
\text{New Row 2} = (1, -2, 3, | 7) - 3(0, 0, 1, | -7) = (1, -2, 0, | -14)
\]

### Final Row Echelon Form:
\[
\begin{pmatrix}
0 & 1 & 0 & | & -19 \\
1 & -2 & 0 & | & -14 \\
0 & 0 & 1 & | & -7
\end{pmatrix}
\]

This is the row echelon form of the system.

---

Would you like further details on any step? Here are some related questions you might find interesting:

1. What is the next step to solve the system from this form?
2. How can row echelon form be used to find solutions to systems of equations?
3. What is the difference between row echelon form and reduced row echelon form?
4. How does Gaussian elimination work in larger systems of equations?
5. Can Gaussian elimination be applied to systems with no solutions?

**Tip**: Row echelon form simplifies solving linear systems, but reduced row echelon form goes further, allowing direct reading of the solution.

turn in row echelon form: −2x1 + 5x2 − 10x3 = −5, x1 − 2x2 + 3x3 = 7, 7x1 − 17x2 + 34x3 = 15

Discover how to apply Gaussian elimination to convert a system of linear equations to row echelon form. This solution includes detailed steps on eliminating variables and matrix manipulation, perfect for undergraduate students studying linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation