线性代数基础线性方程组线性代数做题本(一)解答版选择题不看本题本题讨论已掌握查看解析收入错题#8502已知 η1,η2{\eta }_{1},{\eta }_{2}η1,η2 是非齐次线性方程组 Ax=b{Ax} = bAx=b 的两个不同解, ξ1{\xi }_{1}ξ1, ξ2{\xi }_{2}ξ2 是对应齐次线性方程组 Ax=0{Ax} = 0Ax=0 的基础解系, k1,k2{k}_{1},{k}_{2}k1,k2 为任意常数,则 Ax=b{Ax} = bAx=b 的通解为‾\underline{\qquad}.A. k1ξ1+k2(ξ1+ξ2)+η1−η22{k}_{1}{\xi }_{1} + {k}_{2}\left({{\xi }_{1} + {\xi }_{2}}\right) + \frac{{\eta }_{1} - {\eta }_{2}}{2}k1ξ1+k2(ξ1+ξ2)+2η1−η2B. k1ξ1+k2(ξ1−ξ2)+η1+η22{k}_{1}{\xi }_{1} + {k}_{2}\left({{\xi }_{1} - {\xi }_{2}}\right) + \frac{{\eta }_{1} + {\eta }_{2}}{2}k1ξ1+k2(ξ1−ξ2)+2η1+η2C. k1ξ1+k2(η1+η2)+η1−η22{k}_{1}{\xi }_{1} + {k}_{2}\left({{\eta }_{1} + {\eta }_{2}}\right) + \frac{{\eta }_{1} - {\eta }_{2}}{2}k1ξ1+k2(η1+η2)+2η1−η2D. k1ξ1+k2(η1−η2)+η1+η22{k}_{1}{\xi }_{1} + {k}_{2}\left({{\eta }_{1} - {\eta }_{2}}\right) + \frac{{\eta }_{1} + {\eta }_{2}}{2}k1ξ1+k2(η1−η2)+2η1+η2