Test for consistency of the following system of linear equations and solve
𝑥 + 2𝑦 + 2𝑧 = 1, 2𝑥 + 𝑦 + 𝑧 = 2,3𝑥 + 2𝑦 + 2𝑧 = 3, 𝑦 + 𝑧 = 0

Given the system of linear equations:

1. \( x + 2y + 2z = 1 \)
2. \( 2x + y + z = 2 \)
3. \( 3x + 2y + 2z = 3 \)
4. \( y + z = 0 \)

### Step 1: Set Up the System
To check for consistency, we will use matrix methods to find whether a solution exists. This involves setting up the augmented matrix for the system and using row reduction (Gaussian elimination).

The augmented matrix for this system is:

\[
\begin{bmatrix}
1 & 2 & 2 & | & 1 \\
2 & 1 & 1 & | & 2 \\
3 & 2 & 2 & | & 3 \\
0 & 1 & 1 & | & 0 \\
\end{bmatrix}
\]

### Step 2: Row Reduction
Apply Gaussian elimination to bring this matrix to row echelon form.

1. Subtract 2 times the first row from the second row:
   
   \[
   R_2 = R_2 - 2R_1 \implies \begin{bmatrix} 1 & 2 & 2 & | & 1 \\ 0 & -3 & -3 & | & 0 \\ 3 & 2 & 2 & | & 3 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}
   \]

2. Subtract 3 times the first row from the third row:

   \[
   R_3 = R_3 - 3R_1 \implies \begin{bmatrix} 1 & 2 & 2 & | & 1 \\ 0 & -3 & -3 & | & 0 \\ 0 & -4 & -4 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}
   \]

3. Add the third row to the second row:

   \[
   R_2 = R_2 + R_3 \implies \begin{bmatrix} 1 & 2 & 2 & | & 1 \\ 0 & -7 & -7 & | & 0 \\ 0 & -4 & -4 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}
   \]

4. Multiply the fourth row by 7 and add it to the second row:

   \[
   R_2 = R_2 + 7R_4 \implies \begin{bmatrix} 1 & 2 & 2 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & -4 & -4 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}
   \]

### Step 3: Interpretation
After reducing, the second row in the augmented matrix becomes \( 0 = 0 \), indicating that the system is **consistent**. The matrix has free variables, implying infinitely many solutions.

### Step 4: Solving the System
From the last row:

4. \( y + z = 0 \implies y = -z \).

Substitute \( y = -z \) into the other equations.

From the first equation:
\[
x + 2(-z) + 2z = 1 \implies x = 1.
\]

From the second equation:
\[
2x + y + z = 2 \implies 2(1) + (-z) + z = 2,
\]
which holds for any value of \( z \).

Thus, the solution to the system is:
\[
x = 1, \quad y = -z, \quad z = z \quad (\text{free parameter}).
\]

### Final Solution
The general solution can be written as:
\[
(x, y, z) = (1, -z, z),
\]
where \( z \) is a free parameter.

### Summary
- The system is **consistent**.
- It has infinitely many solutions, parametrized by \( z \).

Would you like more details or have any questions?

### Related Questions
1. How can we use Gaussian elimination to find the rank of a matrix?
2. What conditions must be satisfied for a system of linear equations to have a unique solution?
3. What are free variables in a linear system, and how do they affect the solution set?
4. How can the solution set of a consistent system with infinitely many solutions be expressed in vector form?
5. What is the geometrical interpretation of a system of linear equations with infinitely many solutions?

### Tip
When solving systems of linear equations, always check for the existence of free variables to determine if there are infinitely many solutions.

Test for consistency of the following system of linear equations and solve: x + 2y + 2z = 1, 2x + y + z = 2, 3x + 2y + 2z = 3, y + z = 0

Learn how to test the consistency of a system of linear equations using Gaussian elimination. This example involves setting up an augmented matrix, performing row reduction, and interpreting the solution. Suitable for students in Grades 10-12.

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