System of Equations
You have the following system of equations:
1. \( X + 2Y + 2Z = 2 \)
2. \( 3X - 2Y - Z = 5 \)
3. \( 2X - 5Y + 3Z = -4 \)
4. \( X + 4Y + 6Z = 0 \)

Step 1: Form the Augmented Matrix
The first step is to represent these equations in an augmented matrix form:
\[
\begin{bmatrix}
1 & 2 & 2 & | & 2 \\
3 & -2 & -1 & | & 5 \\
2 & -5 & 3 & | & -4 \\
1 & 4 & 6 & | & 0
\end{bmatrix}
\]

Step 2: Row Operations
Perform row operations to reduce the matrix to row echelon form:
- **R2** = R2 - 3R1
- **R3** = R3 - 2R1
- **R4** = R4 - R1
This gives:
\[
\begin{bmatrix}
1 & 2 & 2 & | & 2 \\
0 & -8 & -7 & | & -1 \\
0 & -9 & -1 & | & -8 \\
0 & 2 & 4 & | & -2
\end{bmatrix}
\]

Step 3: Further Simplification
Continue with row operations to isolate variables:
- Normalize R2 by dividing by -1
- Use R2 to eliminate entries in R3 and R4
Resulting in:
\[
\begin{bmatrix}
1 & 0 & 0 & | & 1 \\
0 & 1 & 1 & | & 1 \\
0 & 0 & 1 & | & -1
\end{bmatrix}
\]

Step 4: Back Substitution
Now, perform back substitution to find the values of \(X, Y, Z\):
- From the last row: \( Z = -1 \)
- Substitute \( Z \) in the second row: \( Y + (-1) = 1 \Rightarrow Y = 2 \)
- Substitute \( Y \) and \( Z \) in the first row: \( X + 2(2) + 2(-1) = 2 \Rightarrow X = 0 \)

Final Solution
Thus, the solution to the system is:
- \( X = 0 \)
- \( Y = 2 \)
- \( Z = -1 \)