19 单选 (4分)已知向量a_(1),a_(2),a_(3),a_(4),a_(5)，且有r(a_(1),a_(2),a_(3),a_(4))=3，r(a_(1),a_(2),a_(3),a_(5))=4，则r(a_(1),a_(2),a_(3),a_(4)+a_(5))=（）A. 1B. 4C. 3D. 2

本题考查向量组的秩的性质及相关计算。解题的关键思路是根据已知向量组的秩判断向量之间的线性关系，再利用向量组秩的性质来求解目标向量组的秩。

步骤一：分析$r(a_{1},a_{2},a_{3},a_{4}) = 3$

向量组的秩是向量组中极大线性无关组所含向量的个数。已知$r(a_{1},a_{2},a_{3},a_{4}) = 3$，这表明向量组$\{a_{1},a_{2},a_{3},a_{4}\}$的极大线性无关组由$3$个向量组成，所以$a_{1},a_{2},a_{3},a_{4}$这$4$个向量是线性相关的，即存在不全为零的实数$k_1,k_2,k_3,k_4$，使得$k_1a_1 + k_2a_2 + k_3a_3 + k_4a_4 = 0$。同时，$a_{1},a_{2},a_{3}$线性无关（因为极大线性无关组有$3$个向量）。

步骤二：分析$r(a_{1},a_{2},a_{3},a_{5}) = 4$

由于$r(a_{1},a_{2},a_{3},a_{5}) = 4$，说明向量组$\{a_{1},a_{2},a_{3},a_{5}\}$的极大线性无关组由$4$个向量组成，所以$a_{1},a_{2},a_{3},a_{5}$这$4$个向量线性无关。

步骤三：判断$a_{4}+a_{5}$与$a_{1},a_{2},a_{3}$的线性关系

假设存在实数$m_1,m_2,m_3$，使得$m_1a_1 + m_2a_2 + m_3a_3 = a_{4}+a_{5}$，移项可得$m_1a_1 + m_2a_2 + m_3a_3 - a_{4}-a_{5} = 0$。
因为$a_{1},a_{2},a_{3},a_{5}$线性无关，若$m_1a_1 + m_2a_2 + m_3a_3 - a_{4}-a_{5} = 0$成立，那么$a_{4}$可由$a_{1},a_{2},a_{3},a_{5}$线性表示，且$a_{1},a_{2},a_{3}$线性无关，这与$a_{1},a_{2},a_{3},a_{4}$线性相关矛盾，所以$a_{4}+a_{5}$不能由$a_{1},a_{2},a_{3}$线性表示。

步骤四：计算$r(a_{1},a_{2},a_{3},a_{4}+a_{5})$

因为$a_{1},a_{2},a_{3}$线性无关，且$a_{4}+a_{5}$不能由$a_{1},a_{2},a_{3}$线性表示，所以向量组$\{a_{1},a_{2},a_{3},a_{4}+a_{5}\}$的极大线性无关组由$4$个向量组成，即$r(a_{1},a_{2},a_{3},a_{4}+a_{5}) = 4$。