Step 1: Express the system of equations in matrix form

We are given the following system of equations:

1. $2x_1 + x_2 + 2x_3 + x_4 = 6$
2. $6x_1 - 6x_2 + 6x_3 + 12x_4 = 36$
3. $4x_1 + 4x_2 + 3x_3 - 3x_4 = -1$
4. $2x_1 + 2x_2 - x_3 + x_4 = 10$

We can write this system in the form $A\mathbf{x} = \mathbf{b}$, where $A$ is the coefficient matrix, $\mathbf{x}$ is the column vector of variables $[x_1, x_2, x_3, x_4]^T$, and $\mathbf{b}$ is the column vector of constants on the right-hand side of the equations.

The coefficient matrix $A$ is:

2 & 1 & 2 & 1 \\ 6 & -6 & 6 & 12 \\ 4 & 4 & 3 & -3 \\ 2 & 2 & -1 & 1 \end{bmatrix}$$ The variable vector $$\mathbf{x}$$ is: $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$$ The constants vector $$\mathbf{b}$$ is: $$\mathbf{b} = \begin{bmatrix} 6 \\ 36 \\ -1 \\ 10 \end{bmatrix}$$ Thus, the system of equations in matrix form is: $$\begin{bmatrix} 2 & 1 & 2 & 1 \\ 6 & -6 & 6 & 12 \\ 4 & 4 & 3 & -3 \\ 2 & 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 6 \\ 36 \\ -1 \\ 10 \end{bmatrix}$$ ### Step 2: Solve the system by the elimination method We now apply the **elimination method** to solve for $$x_1, x_2, x_3, x_4$$. The idea is to manipulate the rows of the augmented matrix to eliminate variables step by step. The augmented matrix corresponding to the system is: $$\left[ \begin{array}{cccc|c} 2 & 1 & 2 & 1 & 6 \\ 6 & -6 & 6 & 12 & 36 \\ 4 & 4 & 3 & -3 & -1 \\ 2 & 2 & -1 & 1 & 10 \\ \end{array} \right]$$ We will perform row operations to reduce this augmented matrix to row echelon form and then solve for the variables. Let me do this and provide you with the steps. sympy as sp # Define the augmented matrix A = sp.Matrix([ [2, 1, 2, 1, 6], [6, -6, 6, 12, 36], [4, 4, 3, -3, -1], [2, 2, -1, 1, 10] ]) # Perform row reduction (Gaussian elimination) A_rref, pivot_columns = A.rref() A_rrefAfter performing the row reduction, we obtain the row echelon form of the augmented matrix: $$\left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{8}{5} \\ 0 & 1 & 0 & 0 & \frac{13}{10} \\ 0 & 0 & 1 & 0 & -\frac{9}{10} \\ 0 & 0 & 0 & 1 & \frac{33}{10} \\ \end{array} \right]$$ From this, we can read off the solutions to the system: $$x_1 = \frac{8}{5}, \quad x_2 = \frac{13}{10}, \quad x_3 = -\frac{9}{10}, \quad x_4 = \frac{33}{10}$$ ### Final Answer: The solution to the system of equations is: $$\boxed{x_1 = \frac{8}{5}, \quad x_2 = \frac{13}{10}, \quad x_3 = -\frac{9}{10}, \quad x_4 = \frac{33}{10}}$$ Would you like any further clarification or additional steps? Here are some related questions to expand your understanding: 1. How do you verify the solution of a system of linear equations? 2. What is the difference between row echelon form and reduced row echelon form? 3. Can Gaussian elimination be used for non-square systems of equations? 4. What is the significance of pivot columns in row reduction? 5. How would the solution change if one of the equations in the system was inconsistent? **Tip:** Always check if the matrix is invertible (i.e., its determinant is non-zero) before solving the system, especially for square systems, as this ensures the system has a unique solution.