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线性代数:求极大无关组的方法: 行阶梯化

例子: $\alpha{1}=(1,1,2,3), \alpha{2}=(1,3,6,1), \alpha{3}=(3,-1, a, 15)$ $\alpha{4}=(1,-5,-10,12)$ $\alpha{1}, \alpha{2}, \alpha{3}, \alpha{4}$线性相关

  1. 求 $a$
  2. 求 $\alpha{1}, \alpha{2}, \alpha{3}, \alpha{4}$ 的一个极大无关组, 并用来表示其余向量.

$\left|\begin{array}{cccc}{1} & {1} & {3} & {1} \ {1} & {3} & {-1} & {-5} \ {2} & {6} & {a} & {-10} \ {3} & {1} & {15} & {12}\end{array}\right| \Rightarrow\left|\begin{array}{cccc}{1} & {1} & {3} & {1} \ {0} & {2} & {-4} & {6} \ {0} & {4} & {a-6} & {-12} \ {0} & {-2} & {6} & {9}\end{array}\right|$ $\Rightarrow 2\left(\begin{array}{cccc}{1} & {1} & {3} & {1} \ {0} & {1} & {-2} & {3} \ {0} & {4} & {a-6} & {-12} \ {0} & {0} & {2} & {3}\end{array}\right)$

$\Rightarrow 2\left|\begin{array}{cccc}{1} & {1} & {3} & {1} \ {0} & {1} & {-2} & {-3} \ {0} & {0} & {a+2} & {0} \ {0} & {0} & {2} & {3}\end{array}\right|$ $\Rightarrow \quad 2 \times 1\times 1\left|\begin{array}{cc}{a+2} & {0} \ {2} & {3}\end{array}\right| \Rightarrow a =-2$

$\Rightarrow\left|\begin{array}{cccc}{1} & {1} & {3} & {1} \ {0} & {1} & {-2} & {-3} \ {0} & {0} & {2} & {3} \ {0} & {0} & {0} & {0}\end{array}\right|$

$\alpha{1} \alpha{2} \alpha_{3}$就是线性无关组

$\alpha{4}-\frac{3}{2} \alpha{3}$ $=\left(\begin{array}{c}{1} \ {-3} \ {3}\end{array}\right)-\left(\begin{array}{c}{\frac{9}{2}} \ {-3} \ {3}\end{array}\right)=-\frac{7}{2} \alpha_{1}$

$\Rightarrow \alpha{4}=\frac{3}{2} \alpha{3}-\frac{7}{2} \alpha_{1}$

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