1. **State the problem:** We have the system of linear equations: $$\frac{1}{4}x - \frac{5}{4}y - \frac{5}{4}z = -10$$ $$\frac{1}{2}x + \frac{3}{2}y + \frac{1}{2}z = 8$$ $$\frac{1}{4}x + \frac{7}{4}y + \frac{3}{4}z = -2$$ We want to solve for $x$, $y$, and $z$ using the inverse of the coefficient matrix. 2. **Write the system in matrix form:** $$A \mathbf{x} = \mathbf{b}$$ where $$A = \begin{bmatrix} \frac{1}{4} & -\frac{5}{4} & -\frac{5}{4} \\ \frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\ \frac{1}{4} & \frac{7}{4} & \frac{3}{4} \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -10 \\ 8 \\ -2 \end{bmatrix}$$ 3. **Formula to solve:** $$\mathbf{x} = A^{-1} \mathbf{b}$$ where $A^{-1}$ is the inverse of matrix $A$. 4. **Find the inverse of $A$:** Calculate $A^{-1}$ (using a calculator or algebraic methods). The inverse is: $$A^{-1} = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 2 \\ 1 & 0 & -2 \end{bmatrix}$$ 5. **Write the solution as a product:** $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 2 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} -10 \\ 8 \\ -2 \end{bmatrix}$$ 6. **Multiply to find the solution:** $$x = 1 \times (-10) + 1 \times 8 + 0 \times (-2) = -10 + 8 + 0 = -2$$ $$y = -1 \times (-10) + 0 \times 8 + 2 \times (-2) = 10 + 0 - 4 = 6$$ $$z = 1 \times (-10) + 0 \times 8 + (-2) \times (-2) = -10 + 0 + 4 = -6$$ **Final solution:** $$x = -2, \quad y = 6, \quad z = -6$$