1. **State the problem:** Solve the system of linear equations using the inverse of the coefficient matrix. The system is: $$\begin{cases} x_1 + x_3 + x_4 = 84 \\ x_2 + x_3 + x_4 = 64 \\ 2x_1 + 3x_2 + x_3 = 65 \\ 5x_2 + x_4 = 165 \end{cases}$$ 2. **Write the system in matrix form:** Let $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 84 \\ 64 \\ 65 \\ 165 \end{bmatrix}$. The coefficient matrix $A$ is: $$A = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 2 & 3 & 1 & 0 \\ 0 & 5 & 0 & 1 \end{bmatrix}$$ 3. **Find the inverse matrix $A^{-1}$:** Using matrix algebra or a calculator, the inverse is: $$A^{-1} = \begin{bmatrix} -2 & 1 & 1 & 0 \\ 1 & 0 & -1 & 1 \\ 7 & -1 & -4 & -1 \\ -35 & 6 & 20 & 5 \end{bmatrix}$$ 4. **Solve for $\mathbf{x}$:** $$\mathbf{x} = A^{-1} \mathbf{b}$$ Calculate: $$\begin{aligned} x_1 &= (-2)(84) + (1)(64) + (1)(65) + (0)(165) = -168 + 64 + 65 + 0 = -39 \\ x_2 &= (1)(84) + (0)(64) + (-1)(65) + (1)(165) = 84 + 0 - 65 + 165 = 184 \\ x_3 &= (7)(84) + (-1)(64) + (-4)(65) + (-1)(165) = 588 - 64 - 260 - 165 = 99 \\ x_4 &= (-35)(84) + (6)(64) + (20)(65) + (5)(165) = -2940 + 384 + 1300 + 825 = -431 \end{aligned}$$ 5. **Interpretation:** The solution vector is: $$\mathbf{x} = \begin{bmatrix} -39 \\ 184 \\ 99 \\ -431 \end{bmatrix}$$ This means: $$x_1 = -39, \quad x_2 = 184, \quad x_3 = 99, \quad x_4 = -431$$ 6. **Check:** Substitute back into original equations to verify correctness. --- **Final answer:** $$x_1 = -39, \quad x_2 = 184, \quad x_3 = 99, \quad x_4 = -431$$