$(a)\;4 \lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\qquad(b)\;\lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\qquad(c)\; \lambda_{1}=2 \lambda_{2}=2 4\lambda_{3}=9\lambda_{4}\qquad(d)\; \lambda_{1}=2 \lambda_{2}=2 3\lambda_{3}=4\lambda_{4}$