4x-2y+z=-5                                                                                                                                                                                                                    5x+y+3z=6                                                                                                                                                                                                     -2x+3y+2z=-4                                                                                                                                                              using Gaussian elimination eliminate z first

To solve this system of linear equations using Gaussian elimination and eliminate \( z \) first, let’s proceed step-by-step. The given system is:

\[
4x - 2y + z = -5 \quad \text{(Equation 1)}
\]
\[
5x + y + 3z = 6 \quad \text{(Equation 2)}
\]
\[
-2x + 3y + 2z = -4 \quad \text{(Equation 3)}
\]

### Step 1: Set up the augmented matrix
We can rewrite the system as an augmented matrix:
\[
\begin{bmatrix}
4 & -2 & 1 & | & -5 \\
5 & 1 & 3 & | & 6 \\
-2 & 3 & 2 & | & -4 \\
\end{bmatrix}
\]

### Step 2: Make the coefficient of \( z \) in the second and third rows zero
To start eliminating \( z \), let's use the first row to eliminate \( z \) from rows 2 and 3.

1. **Eliminate \( z \) from Row 2** by making Row 2 become \( \text{Row 2} - 3 \times \text{Row 1} \):
   \[
   5x + y + 3z - 3(4x - 2y + z) = 6 - 3(-5)
   \]
   Simplifying this:
   \[
   5x + y + 3z - 12x + 6y - 3z = 6 + 15
   \]
   \[
   -7x + 7y + 0z = 21
   \]
   Dividing by 7:
   \[
   -x + y = 3
   \]
   This gives us the updated second row:
   \[
   [ -1, 1, 0 | 3 ]
   \]

2. **Eliminate \( z \) from Row 3** by making Row 3 become \( \text{Row 3} - 2 \times \text{Row 1} \):
   \[
   -2x + 3y + 2z - 2(4x - 2y + z) = -4 - 2(-5)
   \]
   Simplifying this:
   \[
   -2x + 3y + 2z - 8x + 4y - 2z = -4 + 10
   \]
   \[
   -10x + 7y + 0z = 6
   \]
   This gives us the updated third row:
   \[
   [-10, 7, 0 | 6 ]
   \]

The matrix now looks like this:
\[
\begin{bmatrix}
4 & -2 & 1 & | & -5 \\
-1 & 1 & 0 & | & 3 \\
-10 & 7 & 0 & | & 6 \\
\end{bmatrix}
\]

### Step 3: Continue with elimination
We can now continue the Gaussian elimination to get a row-echelon form, allowing us to find \( x \), \( y \), and \( z \).

Would you like to continue with solving for \( x \) and \( y \) as well?

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### Related Questions

1. How can we use back-substitution to solve this system after Gaussian elimination?
2. What are the conditions for a system to have no solution, one solution, or infinitely many solutions?
3. How does Gaussian elimination differ from Gauss-Jordan elimination?
4. How would you handle the system if there was no clear path to eliminate \( z \) directly?
5. How does pivoting help in Gaussian elimination?

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**Tip:** Always keep track of row operations carefully, as a small error can lead to incorrect solutions in Gaussian elimination.

4x - 2y + z = -5, 5x + y + 3z = 6, -2x + 3y + 2z = -4 using Gaussian elimination to eliminate z first

Learn how to solve a system of linear equations using Gaussian elimination with a focus on eliminating the z variable first. This solution includes step-by-step row operations, matrix transformation, and interpretation, covering advanced algebraic concepts for Grades 10-12.

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