线代强化NO9|向量|习题

$$\begin{array}{rl}& \boxed{\alpha },\boxed{\beta },\boxed{\gamma }\text{无关，}\boxed{\alpha },\boxed{\beta }\text{无关，}\boxed{\alpha },\boxed{\beta },\boxed{\delta }\text{相关，故}\boxed{\delta }\text{可由}\boxed{\alpha },\boxed{\beta }\text{线性表示，即}\\ & \boxed{\delta }=k_1\boxed{\alpha }+k_2\boxed{\beta }\\ & \boxed{\delta }\text{也可由}\boxed{\alpha },\boxed{\beta },\boxed{\gamma }\text{线性表示，即}\\ & \boxed{\delta }=k_1\boxed{\alpha }+k_2\boxed{\beta }+0\cdot \boxed{\gamma }\end{array}$$
选C

$$\begin{array}{rl}& \text{解：法1. 观察 法2. 用定义. 设}\boxed{\beta }=k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_m{\boxed{\alpha }}_m\text{，假设}{\boxed{\alpha }}_m\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{表示，即}{\boxed{\alpha }}_m=l_1{\boxed{\alpha }}_1+l_2{\boxed{\alpha }}_2+\cdots +l_{m-1}{\boxed{\alpha }}_{m-1}\\ & \boxed{\beta }=k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_{m-1}{\boxed{\alpha }}_{m-1}+k_m\left(l_1{\boxed{\alpha }}_1+l_2{\boxed{\alpha }}_2+\cdots +l_{m-1}{\boxed{\alpha }}_{m-1}\right)\\ & =\left(k_1+k_m{l}_1\right){\boxed{\alpha }}_1+\left(k_2+k_m{l}_2\right){\boxed{\alpha }}_2+\cdots +\left(k_{m-1}+k_m{l}_{m-1}\right){\boxed{\alpha }}_{m-1}\text{，故}\boxed{\beta }\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{线性表示，}\\ & \text{与已知矛盾. 故}{\boxed{\alpha }}_m\text{不能由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{表示.}\\ & \text{又}\boxed{\beta }=k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_m{\boxed{\alpha }}_m\text{，若}{k}_m=0\text{，则}\boxed{\beta }=k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_{m-1}{\boxed{\alpha }}_{m-1}\\ & \text{故}\boxed{\beta }\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{线性表示，与已知矛盾. 故}{k}_m\ne 0.\\ & {\boxed{\alpha }}_m=-\frac{1}{k_m}\left(k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_{m-1}{\boxed{\alpha }}_{m-1}-\boxed{\beta }\right)\text{，即}{\boxed{\alpha }}_m\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1},\boxed{\beta }\text{表示}\end{array}$$
$$\begin{array}{rl}& \text{法3：用秩. }\boxed{\beta }\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_m\text{表示}⟹r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_m)=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_m\mid \boxed{\beta })\\ & \boxed{\beta }\text{不可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{表示}⟹r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1})<r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\mid \boxed{\beta })\\ & r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1})<r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\mid \boxed{\beta })\le r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_m\mid \boxed{\beta })=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_m)\\ & \text{故}{\boxed{\alpha }}_m\text{不可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\text{表示.}\\ & r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\mid \boxed{\beta })=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1})+1\\ & r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1},{\boxed{\alpha }}_m)=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1})+1⟹r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1}\mid \boxed{\beta })=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1},\boxed{\beta },{\boxed{\alpha }}_m)\\ & \text{故}{\boxed{\alpha }}_m\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_{m-1},\boxed{\beta }\text{线性表示.}\end{array}$$
$$\begin{array}{rl}& \text{【小结】判定、证明抽象向量线性表出的两个基本方法：定义、秩。}\\ & \text{对于定义法：}\\ & ①\text{证明}\boxed{\beta }\text{能由向量组}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m\text{线性表出，需要找到一个包含}\boxed{\beta }\text{和}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m\text{的}\\ & \text{一个等式，再证明}\boxed{\beta }\text{前系数不为}0；\\ & ②\text{证明}\boxed{\beta }\text{不能由向量组}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m\text{线性表出，通常要用反证法。}\\ & \text{通过秩判断：}\\ & ①\boxed{\beta }\text{能由向量组}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m\text{线性表出}⟺r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m)=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m,\boxed{\beta })；\\ & ②\boxed{\beta }\text{不能由向量组}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m\text{线性表出}⟺r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m)<r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\dots ,{\boxed{\alpha }}_m,\boxed{\beta })。\end{array}$$

知识梳理
$$\begin{array}{rl}& ①r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\mid {\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3),{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{线性表示}.\\ & ②r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\mid {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3),{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{可由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{线性表示}.\\ & ③r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\mid {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3),{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{线性表示}.\\ & ④r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\mid {\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3),{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{可由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{线性表示}.\\ & \text{小结：增加一个新的组，秩不变，新的组可由旧的组表示.}\\ & \text{但习惯上把增加的组写在右边.}\end{array}$$
法1
$$\begin{array}{rl}& \text{36. 解：}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{可由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{线性表示}⟹r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\mid {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)\\ & \overline{A}=\left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& \alpha \\ 1& \alpha & \alpha & |& 1& \alpha & 1\\ \alpha & 4& \alpha & |& \alpha & 1& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& \alpha \\ 0& \alpha +2& \alpha +2& |& 0& \alpha -1& 1-\alpha \\ 0& 4+2\alpha & 3\alpha & |& 0& 1-\alpha & 1-{\alpha }^2\end{array}\right)\to \left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& \alpha \\ 0& \alpha +2& \alpha +2& |& 0& \alpha -1& 1-\alpha \\ 0& 0& \alpha -4& |& 0& 3-3\alpha & -(a-1{)}^2\end{array}\right)\\ & ①\text{当}a+2=0\text{时，}a=-2\\ & \overline{A}=\left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& -2\\ 0& 0& 0& |& 0& -3& 3\\ 0& 0& -6& |& 0& 9& -9\end{array}\right)\to \left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& -2\\ 0& 0& -6& |& 0& 9& -9\\ 0& 0& 0& |& 0& -3& 3\end{array}\right)\\ & r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2,r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3,{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3\\ & \text{故}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{不能由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{表示，故}a\ne -2.\end{array}$$
$$\begin{array}{rl}& ②\text{当}a=4\text{时}\\ & \overline{A}=\left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& 4\\ 0& 6& 6& |& 0& 3& -3\\ 0& 0& 0& |& 0& -9& -9\end{array}\right),r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2,r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3,{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3\\ & \text{故}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{不能由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{表示，故}a\ne 4\\ & ③\text{当}a\ne -2\text{且}a\ne 4\text{时},r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=3,r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3,{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3\\ & {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{可由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{线性表示}\end{array}$$
$$\begin{array}{rl}& \overline{B}=({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\mid {\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=\left(\begin{array}{ccccccc}1& 1& \alpha & |& 1& -2& -2\\ 1& \alpha & 1& |& 1& \alpha & \alpha \\ \alpha & 1& 1& |& \alpha & 4& \alpha \end{array}\right)\to \left(\begin{array}{ccccccc}1& 1& \alpha & |& 1& -2& -2\\ 0& \alpha -1& 1-\alpha & |& 0& \alpha +2& \alpha +2\\ 0& 1-\alpha & 1-{\alpha }^2& |& 0& 4+2\alpha & 3\alpha \end{array}\right)\\ & \to \left(\begin{array}{ccccccc}1& 1& \alpha & |& 1& -2& -2\\ 0& \alpha -1& 1-\alpha & |& 0& \alpha +2& \alpha +2\\ 0& 0& (\alpha +2)(1-\alpha )& |& 0& 3\alpha +6& 4\alpha +2\end{array}\right)\\ & \text{若}\alpha =1,\overline{B}=\left(\begin{array}{ccccccc}1& 1& 1& |& 1& -2& -2\\ 0& 0& 0& |& 0& 3& 3\\ 0& 0& 0& |& 0& 9& 6\end{array}\right),\text{则}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{不可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{表示}\\ & \text{若}\alpha \ne 1,\text{则}r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3),{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{表示}\\ & \text{综上，}\alpha =1\end{array}$$
法2
$$\begin{array}{rl}& \text{解：}|{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3|=\left|\begin{array}{ccc}1& 1& \alpha \\ 1& \alpha & 1\\ \alpha & 1& 1\end{array}\right|=-(\alpha -1{)}^2(\alpha +2),\\ & \text{当}\alpha \ne 1\text{且}\alpha \ne -2\text{时，}r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3=r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3),{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{可由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{表示}\\ & \text{当}\alpha =-2\text{时，}({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\mid {\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=\left(\begin{array}{ccccccc}1& 1& -2& |& 1& -2& -2\\ 1& -2& 1& |& 1& -2& -2\\ -2& 1& 1& |& -2& 4& -2\end{array}\right)\to \left(\begin{array}{ccccccc}1& 1& -2& |& 1& -2& -2\\ 0& -3& 3& |& 0& 0& 0\\ 0& 0& 0& |& 0& 0& -6\end{array}\right)\\ & r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=2\ne r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3),{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{不能由}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{表示}\\ & ({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\mid {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=\left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& -2\\ 1& -2& -2& |& 1& -2& 1\\ -2& 4& -2& |& -2& 1& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& -2& -2& |& 1& 1& -2\\ 0& 0& 0& |& 0& -3& 3\\ 0& 0& -6& |& 0& 3& -3\end{array}\right)\end{array}$$
$$\begin{array}{rl}& r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2\ne r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\mid {\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)\text{，故}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{不能由}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{表示}\\ & \text{综上，}\alpha =1\end{array}$$