求解 {ccc}{4}&{-1}&{2}{3}&{2}&{0}{2}&{0}&{1}*{l}{u_{1}}{u_{2}}{u_{3}}=3{l}{u_{1}}{u_{2}}{u_{3}}+{c}{3}{1}{-2}

$$\left( \begin{array} { c c c } { 4 } & { - 1 } & { 2 } \\ { 3 } & { 2 } & { 0 } \\ { 2 } & { 0 } & { 1 } \end{array} \right) \cdot \left( \begin{array} { l } { u _ { 1 } } \\ { u _ { 2 } } \\ { u _ { 3 } } \end{array} \right) = 3 \left( \begin{array} { l } { u _ { 1 } } \\ { u _ { 2 } } \\ { u _ { 3 } } \end{array} \right) + \left( \begin{array} { c } { 3 } \\ { 1 } \\ { - 2 } \end{array} \right)$$

求解 u_1, u_2, u_3 的值

$$u_{1}=u_{3}-1$$

$$u_{2}=3u_{3}-4$$

$$u_{3}\in \mathrm{R}$$