Use gaussian elimination method to solve the system of linear equations: 2x + 3y - z = 5, 4x + 4y - 3z = 3, -2x + 3y-z = 1

To solve the system of linear equations using Gaussian elimination, we follow these steps:

1. Write the augmented matrix for the system.
2. Use row operations to transform the augmented matrix into row-echelon form.
3. Perform back-substitution to find the solution.

Given the system of equations:

\[ \begin{cases} 
2x + 3y - z = 5 \\
4x + 4y - 3z = 3 \\
-2x + 3y - z = 1 
\end{cases} \]

### Step 1: Write the augmented matrix

The augmented matrix for the system is:

\[ \left[\begin{array}{ccc|c}
2 & 3 & -1 & 5 \\
4 & 4 & -3 & 3 \\
-2 & 3 & -1 & 1
\end{array}\right] \]

### Step 2: Use row operations to transform the augmented matrix into row-echelon form

#### First, make the element in the first row, first column a 1 by dividing the entire row by 2:

\[ R1 = \frac{1}{2} R1 \]

\[ \left[\begin{array}{ccc|c}
1 & \frac{3}{2} & -\frac{1}{2} & \frac{5}{2} \\
4 & 4 & -3 & 3 \\
-2 & 3 & -1 & 1
\end{array}\right] \]

#### Next, eliminate the x-term in the second and third rows:

\[ R2 = R2 - 4R1 \]
\[ R3 = R3 + 2R1 \]

\[ \left[\begin{array}{ccc|c}
1 & \frac{3}{2} & -\frac{1}{2} & \frac{5}{2} \\
0 & -2 & -1 & -7 \\
0 & 6 & -2 & 6
\end{array}\right] \]

#### Now, make the element in the second row, second column a 1 by dividing the entire row by -2:

\[ R2 = -\frac{1}{2} R2 \]

\[ \left[\begin{array}{ccc|c}
1 & \frac{3}{2} & -\frac{1}{2} & \frac{5}{2} \\
0 & 1 & \frac{1}{2} & \frac{7}{2} \\
0 & 6 & -2 & 6
\end{array}\right] \]

#### Eliminate the y-term in the first and third rows:

\[ R1 = R1 - \frac{3}{2}R2 \]
\[ R3 = R3 - 6R2 \]

\[ \left[\begin{array}{ccc|c}
1 & 0 & -1 & -5 \\
0 & 1 & \frac{1}{2} & \frac{7}{2} \\
0 & 0 & -5 & -15
\end{array}\right] \]

#### Finally, make the element in the third row, third column a 1 by dividing the entire row by -5:

\[ R3 = -\frac{1}{5} R3 \]

\[ \left[\begin{array}{ccc|c}
1 & 0 & -1 & -5 \\
0 & 1 & \frac{1}{2} & \frac{7}{2} \\
0 & 0 & 1 & 3
\end{array}\right] \]

#### Eliminate the z-term in the first and second rows:

\[ R1 = R1 + R3 \]
\[ R2 = R2 - \frac{1}{2}R3 \]

\[ \left[\begin{array}{ccc|c}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 3
\end{array}\right] \]

### Step 3: Perform back-substitution to find the solution

From the final augmented matrix, we can read off the solutions:

\[ \begin{cases}
x = -2 \\
y = 1 \\
z = 3
\end{cases} \]

Thus, the solution to the system of equations is \( x = -2 \), \( y = 1 \), and \( z = 3 \).

Do you want details or have any questions? Here are some related questions you might find helpful:

1. How does Gaussian elimination work for systems with no solutions or infinite solutions?
2. What are the different types of row operations in Gaussian elimination?
3. How can you verify the solution to a system of linear equations?
4. What is the difference between Gaussian elimination and Gauss-Jordan elimination?
5. How can you apply Gaussian elimination to larger systems of equations?
6. What are the potential pitfalls when using Gaussian elimination?
7. How does the presence of zero rows affect the solution of a system of equations?
8. What is the computational complexity of Gaussian elimination?

**Tip:** Always check the consistency of the system of equations before applying Gaussian elimination to avoid unnecessary computations.

Learn how to solve a system of linear equations using Gaussian elimination with step-by-step instructions. This problem covers fundamental concepts in linear algebra and Gaussian elimination method. Suitable for advanced high school students and college-level learners.

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