Solve {c}{lambda_{1}+lambda_{2}-3lambda_{3}=0}{lambda_{1}-lambda_{2}+5lambda_{3}=0}{lambda_{2}-6lambda_{3}=0}

$$\left\{ \begin{array} { c } { \lambda _ { 1 } + \lambda _ { 2 } - 3 \lambda _ { 3 } = 0 } \\ { \lambda _ { 1 } - \lambda _ { 2 } + 5 \lambda _ { 3 } = 0 } \\ { \lambda _ { 2 } - 6 \lambda _ { 3 } = 0 } \end{array} \right.$$

Solve for λ_1, λ_2, λ_3

$$\lambda _{1}=0$$

$$\lambda _{2}=0$$

$$\lambda _{3}=0$$

Short Steps Using Substitution

Solve \lambda _{1}+\lambda _{2}-3\lambda _{3}=0 for \lambda _{1}.

$$\lambda _{1}=-\lambda _{2}+3\lambda _{3}$$

Substitute -\lambda _{2}+3\lambda _{3} for \lambda _{1} in the equation \lambda _{1}-\lambda _{2}+5\lambda _{3}=0.

$$-\lambda _{2}+3\lambda _{3}-\lambda _{2}+5\lambda _{3}=0$$

Solve the second equation for \lambda _{2} and the third equation for \lambda _{3}.

$$\lambda _{2}=4\lambda _{3} \lambda _{3}=\frac{1}{6}\lambda _{2}$$

Substitute 4\lambda _{3} for \lambda _{2} in the equation \lambda _{3}=\frac{1}{6}\lambda _{2}.

$$\lambda _{3}=\frac{1}{6}\times 4\lambda _{3}$$

Solve \lambda _{3}=\frac{1}{6}\times 4\lambda _{3} for \lambda _{3}.

$$\lambda _{3}=0$$

Substitute 0 for \lambda _{3} in the equation \lambda _{2}=4\lambda _{3}.

$$\lambda _{2}=4\times 0$$

Calculate \lambda _{2} from \lambda _{2}=4\times 0.

$$\lambda _{2}=0$$

Substitute 0 for \lambda _{2} and 0 for \lambda _{3} in the equation \lambda _{1}=-\lambda _{2}+3\lambda _{3}.

$$\lambda _{1}=-0+3\times 0$$

Calculate \lambda _{1} from \lambda _{1}=-0+3\times 0.

$$\lambda _{1}=0$$

The system is now solved.

$$\lambda _{1}=0 \lambda _{2}=0 \lambda _{3}=0$$