题目
【例19】(2025-2)设矩阵A=(alpha_(1),alpha_(2),alpha_(3),alpha_(4)).若alpha_(1),alpha_(2),alpha_(3)线性无关，且alpha_(1)+alpha_(2)=alpha_(3)+alpha_(4)，则方程组AX=alpha_(1)+4alpha_(4)的通解为X=____.
【例19】(2025-2)设矩阵$A=(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})$.若$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关，且$\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}$，则方程组$AX=\alpha_{1}+4\alpha_{4}$的通解为X=____.
题目解答
答案
已知 $\alpha_1, \alpha_2, \alpha_3$ 线性无关，且 $\alpha_4 = \alpha_1 + \alpha_2 - \alpha_3$，则矩阵 $A$ 的秩为 3。齐次方程组 $AX = 0$ 的基础解系为 $\eta = (1, 1, -1, -1)^T$。
将 $\alpha_4$ 代入非齐次方程组，得
\[
\alpha_1 + 4\alpha_4 = 5\alpha_1 + 4\alpha_2 - 4\alpha_3,
\]
故特解为 $\xi = (5, 4, -4, 0)^T$。
通解为特解加齐次解，即
\[
\boxed{(5, 4, -4, 0)^T + k(1, 1, -1, -1)^T}.
\]