1. **State the problem:** Solve the system of equations by calculating the inverse of the coefficient matrix using elementary row operations. The system is: $$\begin{cases} x_1 + x_2 + x_3 + x_4 = 0 \\ x_1 + x_2 + x_3 - x_4 = 4 \\ x_1 + x_2 - x_3 + x_4 = -4 \\ x_1 - x_2 + x_3 + x_4 = 2 \end{cases}$$ 2. **Write the system in matrix form:** $$A \mathbf{x} = \mathbf{b}$$ where $$A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 1 & -1 & 1 & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \\ 4 \\ -4 \\ 2 \end{bmatrix}$$ 3. **Find the inverse of matrix $A$ using elementary row operations:** Augment $A$ with the identity matrix $I$: $$[A | I] = \left[ \begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & -1 & 0 & 1 & 0 & 0 \\ 1 & 1 & -1 & 1 & 0 & 0 & 1 & 0 \\ 1 & -1 & 1 & 1 & 0 & 0 & 0 & 1 \end{array} \right]$$ Perform row operations to convert the left side to the identity matrix: - Subtract row 1 from rows 2, 3, and 4: $$R_2 = R_2 - R_1, R_3 = R_3 - R_1, R_4 = R_4 - R_1$$ Result: $$\left[ \begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & -1 & 1 & 0 & 0 \\ 0 & 0 & -2 & 0 & -1 & 0 & 1 & 0 \\ 0 & -2 & 0 & 0 & -1 & 0 & 0 & 1 \end{array} \right]$$ - Divide rows 2, 3, and 4 by their leading coefficients: $$R_2 = -\frac{1}{2} R_2, R_3 = -\frac{1}{2} R_3, R_4 = -\frac{1}{2} R_4$$ Result: $$\left[ \begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{2} & 0 & 0 & -\frac{1}{2} \end{array} \right]$$ - Use rows 2, 3, and 4 to eliminate the corresponding entries in row 1: $$R_1 = R_1 - R_2 - R_3 - R_4$$ Calculate: Left side becomes $[1,0,0,0]$ and right side: $$1 - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{1}{2}, \quad 0 + \frac{1}{2} - 0 - 0 = \frac{1}{2}, \quad 0 - 0 + \frac{1}{2} - 0 = \frac{1}{2}, \quad 0 - 0 - 0 + \frac{1}{2} = \frac{1}{2}$$ So row 1 becomes: $$[1,0,0,0 | -\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}]$$ - The matrix is now: $$\left[ \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{2} & 0 & 0 & -\frac{1}{2} \end{array} \right]$$ - Swap rows 2 and 4 to get the identity matrix in order: $$\left[ \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & 0 & \frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 0 \end{array} \right]$$ Thus, $$A^{-1} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{2} & 0 & 0 \end{bmatrix}$$ 4. **Calculate the solution vector:** $$\mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ \frac{1}{2} & 0 & -\frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{2} & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 4 \\ -4 \\ 2 \end{bmatrix}$$ Calculate each component: - $x_1 = -\frac{1}{2} \cdot 0 + \frac{1}{2} \cdot 4 + \frac{1}{2} \cdot (-4) + \frac{1}{2} \cdot 2 = 0 + 2 - 2 + 1 = 1$ - $x_2 = \frac{1}{2} \cdot 0 + 0 \cdot 4 + 0 \cdot (-4) - \frac{1}{2} \cdot 2 = 0 + 0 + 0 - 1 = -1$ - $x_3 = \frac{1}{2} \cdot 0 + 0 \cdot 4 - \frac{1}{2} \cdot (-4) + 0 \cdot 2 = 0 + 0 + 2 + 0 = 2$ - $x_4 = \frac{1}{2} \cdot 0 - \frac{1}{2} \cdot 4 + 0 \cdot (-4) + 0 \cdot 2 = 0 - 2 + 0 + 0 = -2$ 5. **Final answer:** $$\boxed{\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 2 \\ -2 \end{bmatrix}}$$