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:

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.