线代强化NO10|向量|习题

$$\begin{array}{rl}& \text{解：}({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\mid {\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=\left(\begin{array}{ccccccc}1& 1& 1& |& 1& 0& 1\\ 1& 0& 2& |& 1& 2& 3\\ 4& 4& a^2+3& |& a+3& 1-a& a^2+3\end{array}\right)\to \left(\begin{array}{ccccccc}1& 1& 1& |& 1& 0& 1\\ 0& -1& 1& |& 0& 2& 2\\ 0& 0& a^2-1& |& a-1& 1-a& a^2-1\end{array}\right)\\ & ({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=\left(\begin{array}{ccc}1& 0& 1\\ 1& 2& 3\\ a+3& 1-a& a^2+3\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 2\\ 0& 1-a& a^2-a\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& a^2-1\end{array}\right)\\ & ①\text{当}a\ne \pm 1\text{时}\\ & \text{可以读出}r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=3,r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=3,r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=3\\ & \text{则}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{与}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{等价}\end{array}$$
$$\begin{array}{rl}& ②\text{当}a=1\text{时，}r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2,r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=2,r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2\\ & \text{则}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{与}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{等价}\\ & ③\text{当}a=-1\text{时，}r({\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=2,r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)=2,r({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3,{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3)=3\\ & \text{则}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{与}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{不等价}\\ & \text{故}a\ne -1\end{array}$$
$$\begin{array}{rl}& \text{故}a\ne -1\\ & \text{当}a=1\text{时，}\left(\begin{array}{ccccc}1& 1& 1& |& 1\\ 0& -1& 1& |& 2\\ 0& 0& 0& |& 0\end{array}\right)\to \left(\begin{array}{ccccc}1& 0& 2& |& 3\\ 0& 1& -1& |& -2\\ 0& 0& 0& |& 0\end{array}\right),\text{故}{\boxed{\beta }}_3=(3-2k){\boxed{\alpha }}_1+(-2+k){\boxed{\alpha }}_2+k{\boxed{\alpha }}_3,k\in \mathbb{R}\\ & \text{当}a\ne \pm 1\text{时，}\left(\begin{array}{ccccc}1& 1& 1& |& 1\\ 0& -1& 1& |& 2\\ 0& 0& a^2-1& |& a^2-1\end{array}\right)\to \left(\begin{array}{ccccc}1& 1& 1& |& 1\\ 0& 1& -1& |& -2\\ 0& 0& 1& |& 1\end{array}\right)\to \left(\begin{array}{ccccc}1& 0& 0& |& 1\\ 0& 1& 0& |& -1\\ 0& 0& 1& |& 1\end{array}\right),{\boxed{\beta }}_3={\boxed{\alpha }}_1-{\boxed{\alpha }}_2+{\boxed{\alpha }}_3\end{array}$$
$$\begin{array}{rl}& \text{【小结】}\\ & 1.\text{向量组}{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n\text{能由向量组}{\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m\text{线性表出，相当于线性方程组}\\ & ({\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m)\boxed{x}={\boxed{\alpha }}_i(i=1,\cdots ,n)\text{都有解。由于这}n\text{个线性方程组的系数矩阵一样（都为}\\ & ({\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m)\text{），所以我们可以将这}n\text{个线性方程组写在一起，组成一个大型增广矩阵}\\ & ({\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m,{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n)\text{，再做初等行变换。}\\ & 2.\text{向量组}{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n\text{中只要有一个向量无法被另一个向量组}{\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m\text{线性表出，就称为}\\ & \text{向量组}{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n\text{不能由向量组}{\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m\text{线性表出。}\\ & 3.{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n\text{与}{\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m\text{等价}\iff r({\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n)=r({\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m)=r({\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n,{\boxed{\beta }}_1,\cdots ,{\boxed{\beta }}_m)。\end{array}$$

线性相关与线性无关的判定与证明

$$\begin{array}{rl}& \text{解：法1：}\\ & \left(\begin{array}{cccc}2& 2& 3& 4\\ 1& 1& 2& 3\\ 1& a& 1& 2\\ 1& a& a& 1\end{array}\right)\to \left(\begin{array}{cccc}1& 1& 2& 3\\ 0& 0& -1& -2\\ 0& a-1& -1& -1\\ 0& a-1& a-2& -2\end{array}\right)\to \left(\begin{array}{cccc}1& 1& 2& 3\\ 0& a-1& -1& -1\\ 0& 0& a-1& -1\\ 0& 0& -1& -2\end{array}\right)\to \left(\begin{array}{cccc}1& 1& 2& 3\\ 0& a-1& -1& -1\\ 0& 0& -1& -2\\ 0& 0& 0& 1-2a\end{array}\right)\\ & \text{因向量组线性相关且}a\ne 1\text{，故}1-2a=0⟹a=\frac{1}{2}\\ & \text{法2：}\\ & \left|\begin{array}{cccc}2& 2& 3& 4\\ 1& 1& 2& 3\\ 1& a& 1& 2\\ 1& a& a& 1\end{array}\right|=\left|\begin{array}{cccc}0& 0& -1& -2\\ 1& 1& 2& 3\\ 0& a-1& -1& -1\\ 0& a-1& a-2& -2\end{array}\right|=1\cdot (-1{)}^{1+2}\left|\begin{array}{ccc}0& -1& -2\\ a-1& -1& -1\\ a-1& a-2& -2\end{array}\right|=\left|\begin{array}{ccc}0& 1& 2\\ a-1& -1& -1\\ 0& a-1& -1\end{array}\right|\\ & =(-1{)}^{1+2}(a-1)\left|\begin{array}{cc}1& 2\\ a-1& -1\end{array}\right|=-(a-1)(1-2a)=0\\ & \text{因}a\ne 1\text{，故}1-2a=0⟹a=\frac{1}{2}\end{array}$$

$$\left(\begin{array}{cccc}0& 0& 1& -1\\ 0& 1& -1& 1\\ c_1& c_2& c_3& c_4\end{array}\right)\to \left(\begin{array}{cccc}0& 0& 1& -1\\ 0& 1& 0& 0\\ c_1& c_2& c_3& c_4\end{array}\right)$$
选C
$$\begin{array}{rl}& \text{【小结】}\\ & 1.\text{数值型向量组的线性相关性基本思路是通过秩来判断：满秩，无关；不满秩，相关；}\\ & 2.\text{如果}{\boxed{\alpha }}_1,\cdots ,{\boxed{\alpha }}_n\text{恰好为}n\text{维向量组，可以组成方阵，就考虑通过行列式来判断：行列式不为}0\text{，无关；为}0\text{，相关。}\end{array}$$

$$\begin{array}{rl}& \text{解：}({\boxed{\alpha }}_1-{\boxed{\alpha }}_2,{\boxed{\alpha }}_2-{\boxed{\alpha }}_3,{\boxed{\alpha }}_3-{\boxed{\alpha }}_1)=({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)\left(\begin{array}{ccc}1& 0& -1\\ -1& 1& 0\\ 0& -1& 1\end{array}\right)\\ & \left|\begin{array}{ccc}1& 0& -1\\ -1& 1& 0\\ 0& -1& 1\end{array}\right|=\left|\begin{array}{ccc}1& 0& -1\\ 0& 1& -1\\ 0& -1& 1\end{array}\right|=0,r({\boxed{\alpha }}_1-{\boxed{\alpha }}_2,{\boxed{\alpha }}_2-{\boxed{\alpha }}_3,{\boxed{\alpha }}_3-{\boxed{\alpha }}_1)=r\left(\begin{array}{ccc}1& 0& -1\\ -1& 1& 0\\ 0& -1& 1\end{array}\right)<3\\ & \text{故}{\boxed{\alpha }}_1-{\boxed{\alpha }}_2,{\boxed{\alpha }}_2-{\boxed{\alpha }}_3,{\boxed{\alpha }}_3-{\boxed{\alpha }}_1\text{相关}\end{array}$$
$$\begin{array}{rl}& \text{【小结】}\\ & \text{若}n\text{维向量}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{线性无关，且满足}\\ & [{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3]=[{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3]C,\\ & \text{则}{\boxed{\beta }}_1,{\boxed{\beta }}_2,{\boxed{\beta }}_3\text{线性无关的充要条件是行列式}|C|\ne 0.\end{array}$$

$$\begin{array}{rl}& \text{解：}({\boxed{\alpha }}_1+k{\boxed{\alpha }}_3,{\boxed{\alpha }}_2+l{\boxed{\alpha }}_3)=({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)\left(\begin{array}{cc}1& 0\\ 0& 1\\ k& l\end{array}\right)\\ & \text{若}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{无关，}({\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3)\text{可逆，}r({\boxed{\alpha }}_1+k{\boxed{\alpha }}_3,{\boxed{\alpha }}_2+l{\boxed{\alpha }}_3)=r\left(\begin{array}{cc}1& 0\\ 0& 1\\ k& l\end{array}\right)=2,\text{则}{\boxed{\alpha }}_1+k{\boxed{\alpha }}_3,{\boxed{\alpha }}_2+l{\boxed{\alpha }}_3\text{无关}\\ & \text{若}{\boxed{\alpha }}_1+k{\boxed{\alpha }}_3,{\boxed{\alpha }}_2+l{\boxed{\alpha }}_3\text{无关，}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{不一定无关}\\ & \text{反例：令}{\boxed{\alpha }}_3=0,{\boxed{\alpha }}_1,{\boxed{\alpha }}_2\text{无关，但}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,{\boxed{\alpha }}_3\text{相关}\\ & A\Rightarrow B\\ & A\text{是}B\text{的充分}\\ & B\text{是}A\text{的必要}\\ & B\text{的充分是}A\\ & A\text{的必要是}B\end{array}$$
选A

选B
存在一组不全为0的数

$$\begin{array}{rl}& \{\begin{array}{llc}{a}_{11}{b}_1+a_{12}{b}_2+\cdots +a_{1n}{b}_n=0& {\boxed{\alpha }}_1^T\boxed{\beta }=0& {\boxed{\beta }}^T{\boxed{\alpha }}_1=0\\ a_{21}{b}_1+a_{22}{b}_2+\cdots +a_{2n}{b}_n=0& {\boxed{\alpha }}_2^T\boxed{\beta }=0& {\boxed{\beta }}^T{\boxed{\alpha }}_2=0\\ ⋮& ⋮& ⋮\\ a_{r1}{b}_1+a_{r2}{b}_2+\cdots +a_{rn}{b}_n=0& {\boxed{\alpha }}_r^T\boxed{\beta }=0& {\boxed{\beta }}^T{\boxed{\alpha }}_r=0\end{array}\\ & \text{解：由}\boxed{\beta }\text{为方程的解可知，}\\ & {\boxed{\alpha }}_1^T\boxed{\beta }=0,{\boxed{\alpha }}_2^T\boxed{\beta }=0,\cdots ,{\boxed{\alpha }}_r^T\boxed{\beta }=0,\text{即}{\boxed{\beta }}^T{\boxed{\alpha }}_1={\boxed{\beta }}^T{\boxed{\alpha }}_2=\cdots ={\boxed{\beta }}^T{\boxed{\alpha }}_r=0\\ & \text{设}{k}_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_r{\boxed{\alpha }}_r+k_{r+1}\boxed{\beta }=0\\ & k_1{\boxed{\beta }}^T{\boxed{\alpha }}_1+k_2{\boxed{\beta }}^T{\boxed{\alpha }}_2+\cdots +k_r{\boxed{\beta }}^T{\boxed{\alpha }}_r+k_{r+1}{\boxed{\beta }}^T\boxed{\beta }=k_{r+1}{\boxed{\beta }}^T\boxed{\beta }=k_{r+1}(b_1^2+b_2^2+\cdots +b_n^2)=0\\ & \text{则}{k}_{r+1}=0.\text{又}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_r\text{无关，故}{k}_1=k_2=\cdots =k_r=0\end{array}$$
$$\text{故}{\boxed{\alpha }}_1,{\boxed{\alpha }}_2,\cdots ,{\boxed{\alpha }}_r,\boxed{\beta }\text{无关}.$$

$$\begin{array}{rl}& \text{【小结】证明向量组线性无关最常用的方法是利用定义：先假设}\\ & k_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_n{\boxed{\alpha }}_n=\boxed{0},\\ & \text{再对该等式进行恒等变形，借助于已知条件说明其系数}{k}_1,k_2,\cdots ,k_n\text{全为}0\text{。}\\ & \text{此时常用的思路有两种：}\\ & \text{一是将式}{k}_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_n{\boxed{\alpha }}_n=\boxed{0}\text{展开，化为关于}{k}_1,k_2,\cdots ,k_n\text{的齐次线性方程组；}\\ & \text{另一种思路是按照已知的信息，在等式两边同时乘以某矩阵之后对等式进行重组，}\\ & \text{如果处理得当，运算之后等式}{k}_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_n{\boxed{\alpha }}_n=\boxed{0}\text{将会得到简化，}\\ & \text{从而利用条件推导出系数}{k}_1,k_2,\cdots ,k_n\text{全为}0\text{。}\\ & \text{其中后一种方法适用范围较广，考试对它的考查也较多，使用它的关键是选择等式}\\ & \text{两边要乘的矩阵，这里的基本原则是尽量向已知条件靠拢，尽量简化等式}{k}_1{\boxed{\alpha }}_1+k_2{\boxed{\alpha }}_2+\cdots +k_n{\boxed{\alpha }}_n=\boxed{0}\text{。}\end{array}$$

选A
$${\beta }_1=k_1{\alpha }_1+k_2{\alpha }_2+k_3{\alpha }_3$$

选A

$$\begin{array}{rl}& \text{解：设}{\alpha }_1\alpha +{\alpha }_{2}A\alpha +\cdots +{\alpha }_k{A}^{k-1}\alpha =0\\ & \text{同时乘}{A}^{k-1},{\alpha }_1{A}^{k-1}\alpha +{\alpha }_2{A}^k\alpha +\cdots +{\alpha }_k{A}^{2k-2}\alpha =0\\ & {\alpha }_1{A}^{k-1}\alpha =0,{\alpha }_1=0,\text{依次类推，可证明}{\alpha }_1={\alpha }_2=\cdots ={\alpha }_k=0\\ & \text{则}\alpha ,A\alpha ,\cdots ,A^{k-1}\alpha \text{线性无关}\end{array}$$

$$\begin{array}{rl}& \text{解：设}B=({\boxed{\beta }}_1,{\boxed{\beta }}_2,\cdots ,{\boxed{\beta }}_n)\text{，则}\\ & x_1{\boxed{\beta }}_1+x_2{\boxed{\beta }}_2+\cdots +x_n{\boxed{\beta }}_n=0,({\boxed{\beta }}_1,{\boxed{\beta }}_2,\cdots ,{\boxed{\beta }}_n)\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)=0,Bx=0\\ & \text{左乘}A,ABx=0,Ex=0,x=0,Bx=0\text{只有零解}\\ & \text{故}B\text{的列向量组线性无关}\end{array}$$
$$\begin{array}{rl}& \text{改：}\\ & \text{①}AB=C,C\text{可逆}⟹B\text{的列向量组线性无关}\\ & \text{②}AB=C,C\text{不可逆}⟹B\text{的列向量组无法判定}\\ & \text{若}B=0,B\text{的列向量组线性相关}\\ & \text{若}A=\left(\begin{array}{ccc}1& 1& 0\\ 1& 2& 0\end{array}\right),B=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right),C=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right),B\text{的列向量组线性无关}\end{array}$$