1. **Stating the problem:** Given constants $K_1=1$, $M_1=1$, $X_1(0)=1$, $K_2=2$, $M_2=1$, $X_2(0)=0$, and $K_3=2$, we analyze the system of equations involving $X_1$ and $X_2$. 2. **Formulating the equations:** From the problem, the forces and masses relate as: $$M_1 = -K_1(X_1 - 0) - K_2(X_1 - 0) - K_3(X_1 - X_2)$$ Simplifying the force $F_1$: $$F_1 = -K_1 X_1 - K_2 X_1 - K_3 X_1 + K_3 X_2 = X_1(K_1 + K_2 + K_3) - X_2 K_3$$ Similarly for $M_2$: $$M_2 = -K_3 (X_2 - X_1)$$ which gives $$F_2 = K_3 X_2 - K_3 X_1$$ 3. **Matrix form of the system:** The system can be written as: $$\begin{pmatrix} K_1 + K_2 + K_3 & -K_3 \\ -K_3 & K_3 \end{pmatrix} = \begin{pmatrix} 1+2+2 & -2 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix}$$ 4. **Finding eigenvalues $\Lambda$:** We solve the characteristic equation: $$\det\left( \begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix} - \begin{pmatrix} \Lambda & 0 \\ 0 & \Lambda \end{pmatrix} \right) = 0$$ which is: $$\det\begin{pmatrix} 5 - \Lambda & -2 \\ -2 & 2 - \Lambda \end{pmatrix} = (5 - \Lambda)(2 - \Lambda) - (-2)(-2) = 0$$ 5. **Expanding and simplifying:** $$ (5 - \Lambda)(2 - \Lambda) - 4 = 10 - 5\Lambda - 2\Lambda + \Lambda^2 - 4 = \Lambda^2 - 7\Lambda + 6 = 0 $$ 6. **Factoring the quadratic:** $$ (\Lambda - 6)(\Lambda - 1) = 0 $$ 7. **Eigenvalues:** $$ \Lambda = 6 \quad \text{or} \quad \Lambda = 1 $$ **Final answer:** The eigenvalues of the system matrix are $\boxed{6}$ and $\boxed{1}$.