{l}{2x_{1}-2x_{2}+x_{3}=lambdax_{1}}{2x_{1}-3x_{2}+2x_{3}=lambdax_{2}}{-x_{1}+2x_{2}=lambdax_{2}} を解きます| Microsoft 数学ソルバー

$$\left. \begin{array} { l } { 2 x _ { 1 } - 2 x _ { 2 } + x _ { 3 } = \lambda x _ { 1 } } \\ { 2 x _ { 1 } - 3 x _ { 2 } + 2 x _ { 3 } = \lambda x _ { 2 } } \\ { - x _ { 1 } + 2 x _ { 2 } = \lambda x _ { 2 } } \end{array} \right.$$

x_1,x_2,x_3 を解く (複素数の解)

$$\left\{\begin{matrix}\\x_{1}=0\text{, }x_{2}=0\text{, }x_{3}=0\text{, }&\text{unconditionally}\\x_{1}=x_{2}\left(2-\lambda \right)\text{, }x_{2}\in \mathrm{C}\text{, }x_{3}=-x_{2}\left(\lambda ^{2}-4\lambda +2\right)\text{, }&\lambda =1\text{ or }\lambda =\frac{3}{2}\end{matrix}\right.$$

x_1,x_2,x_3 を解く

$$\left\{\begin{matrix}x_{1}=0\text{, }x_{2}=0\text{, }x_{3}=0\text{, }&\lambda \neq 1\text{ and }\lambda \neq \frac{3}{2}\\x_{1}=x_{2}\left(2-\lambda \right)\text{, }x_{2}\in \mathrm{R}\text{, }x_{3}=-x_{2}\left(\lambda ^{2}-4\lambda +2\right)\text{, }&\lambda =1\text{ or }\lambda =\frac{3}{2}\end{matrix}\right.$$