Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION.)
x1	 + 	2x2	 − 	3x3	 = 	19
2x1	 − 	x2	 + 	x3	 = 	0
4x1	 − 	x2	 + 	x3	 = 	8

x1	

x2
x3
=	


We can solve the given system of linear equations using the **Gauss-Jordan elimination** method. The system is:

\[
x_1 + 2x_2 - 3x_3 = 19
\]
\[
2x_1 - x_2 + x_3 = 0
\]
\[
4x_1 - x_2 + x_3 = 8
\]

We will express this system as an augmented matrix and perform row operations to reduce it to reduced row echelon form (RREF).

### Step 1: Write the augmented matrix
The augmented matrix for this system is:

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
2 & -1 & 1 & | & 0 \\
4 & -1 & 1 & | & 8
\end{pmatrix}
\]

### Step 2: Perform row operations to reduce the matrix

1. **Row 2 operation**: \( R_2 \leftarrow R_2 - 2R_1 \)

This will make the leading coefficient of the second row zero.

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & -5 & 7 & | & -38 \\
4 & -1 & 1 & | & 8
\end{pmatrix}
\]

2. **Row 3 operation**: \( R_3 \leftarrow R_3 - 4R_1 \)

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & -5 & 7 & | & -38 \\
0 & -9 & 13 & | & -68
\end{pmatrix}
\]

3. **Row 3 operation**: \( R_3 \leftarrow R_3 - \left( \frac{9}{5} \right) R_2 \)

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & -5 & 7 & | & -38 \\
0 & 0 & 0.4 & | & 2.4
\end{pmatrix}
\]

4. **Row 3 operation**: Multiply \( R_3 \) by \( \frac{5}{2} \) to simplify the third row.

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & -5 & 7 & | & -38 \\
0 & 0 & 1 & | & 6
\end{pmatrix}
\]

5. **Row 2 operation**: Eliminate the third element of \( R_2 \), \( R_2 \leftarrow R_2 - 7R_3 \).

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & -5 & 0 & | & -80 \\
0 & 0 & 1 & | & 6
\end{pmatrix}
\]

6. **Row 2 operation**: Multiply \( R_2 \) by \( \frac{-1}{5} \) to simplify.

\[
\begin{pmatrix}
1 & 2 & -3 & | & 19 \\
0 & 1 & 0 & | & 16 \\
0 & 0 & 1 & | & 6
\end{pmatrix}
\]

7. **Row 1 operation**: Eliminate the second element of \( R_1 \), \( R_1 \leftarrow R_1 - 2R_2 \).

\[
\begin{pmatrix}
1 & 0 & -3 & | & -13 \\
0 & 1 & 0 & | & 16 \\
0 & 0 & 1 & | & 6
\end{pmatrix}
\]

8. **Row 1 operation**: Eliminate the third element of \( R_1 \), \( R_1 \leftarrow R_1 + 3R_3 \).

\[
\begin{pmatrix}
1 & 0 & 0 & | & 5 \\
0 & 1 & 0 & | & 16 \\
0 & 0 & 1 & | & 6
\end{pmatrix}
\]

### Final Solution:
\[
x_1 = 5, \quad x_2 = 16, \quad x_3 = 6
\]

Thus, the solution to the system is:
\[
x_1 = 5, \quad x_2 = 16, \quad x_3 = 6
\]

Would you like more details on any step?

### Related Questions:
1. How is the Gauss-Jordan elimination method different from Gaussian elimination?
2. Can you explain the conditions for a system of equations to have no solution?
3. How do row operations affect the solutions of a system?
4. What are the potential outcomes of solving a system of equations?
5. How do you verify the solution of a system of linear equations?

**Tip:** Always check the determinant of the coefficient matrix for square systems; if it’s zero, the system may not have a unique solution.

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION.)

x1 + 2x2 − 3x3 = 19
2x1 − x2 + x3 = 0
4x1 − x2 + x3 = 8

Learn how to solve the system of linear equations: x1 + 2x2 - 3x3 = 19, 2x1 - x2 + x3 = 0, and 4x1 - x2 + x3 = 8 using Gauss-Jordan elimination. This step-by-step guide explains key linear algebra concepts, including row operations and matrix transformations. Suitable for college-level students studying Linear Algebra.

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