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向量組、方程組與線性空間

由基礎(chǔ)解系反解方程組

(東華大學(xué),2021)已知向量組
${ \alpha_{1}=(1,3,-2,2,0)^{\prime},\alpha_{2}=(1,-3,2,0,4)^{\prime},\alpha_{3}=(3,3,-2,4,4)^{\prime} . }$
記 ${M=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ 為 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 生成的子空間.

求一個(gè)以 ${M}$ 為解空間的齊次線性方程組 (I);

求一個(gè)導(dǎo)出組為 (I),有一個(gè)特解為 ${\alpha_{0}=(1,-3,3,0,0)^{\prime}}$ 的非齊次線性方程組 (II)

solution

首先記 ${A=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,對(duì) ${A^{\prime}}$ 進(jìn)行初等行變換,化為階梯形,有
$A^{\prime}=\left(\begin{array}{ccccc} 1 & 3 & -2 & 2 & 0 \\ 1 & -3 & 2 & 0 & 4 \\ 3 & 3 & -2 & 4 & 4 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -2 & 2 & 0 \\ 0 & -6 & 4 & -2 & 4 \\ 0 & -6 & 4 & -2 & 4 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 0 & 0 & 1 & 2 \\ 0 & 3 & -2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$
由此可知 ${r(A)=2}$ ,而明顯 ${\alpha_{1},\alpha_{2}}$ 線性無(wú)關(guān),所以 ${\alpha_{1},\alpha_{2}}$ 為 ${M}$ 的一組基.另外,根據(jù)上述階梯形可知方程組 ${A^{\prime} X=0}$ 的基礎(chǔ)解系為
${ \eta_{1}=(0,2,3,0,0)^{\prime},\eta_{2}=(-3,-1,0,3,0)^{\prime},\eta_{3}=(-6,2,0,0,3)^{\prime} }$
記 ${B=\left(\eta_{1},\eta_{2},\eta_{3}\right)}$ ,則有 ${A^{\prime} B=O}$ ,從而 ${B^{\prime} A=O}$ ,這說(shuō)明 ${A}$ 的列向量 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 均為方程組 ${B^{\prime} X=0}$ 的解,而明顯 ${r(B)=3}$ ,所以方程組 ${B^{\prime} X=0}$ 的基礎(chǔ)解系當(dāng)中含有 ${5-3=2}$ 個(gè)向量,而 ${\alpha_{1},\alpha_{2}}$ 線性無(wú)關(guān),所以它們構(gòu)成 ${B^{\prime} X=0}$ 的基礎(chǔ)解系,也就是說(shuō)方程組 ${(I): B^{\prime} X=0}$ ,即
${ (I)\left\{\begin{array}{l} 2 x_{2}+3 x_{3}=0 \\ -3 x_{1}-x_{2}+3 x_{4}=0 \\ -6 x_{1}+2 x_{2}+3 x_{5}=0 \end{array}\right. }$
的解空間為 ${M}$ .
當(dāng) ${x_{1}=1,x_{2}=-3,x_{3}=3,x_{4}=x_{5}=0}$ 時(shí),有
${ 2 x_{2}+3 x_{3}=3 ; -3 x_{1}-x_{2}+3 x_{4}=0 ; -6 x_{1}+2 x_{2}+3 x_{5}=-12 . }$
于是方程組
${ (I I)\left\{\begin{array}{l} 2 x_{2}+3 x_{3}=3 \\ -3 x_{1}-x_{2}+3 x_{4}=0 \\ -6 x_{1}+2 x_{2}+3 x_{5}=-12 . \end{array}\right. }$
的導(dǎo)出組為 ${(I)}$ ,同時(shí)以 ${\alpha_{0}=(1,-3,3,0,0)^{\prime}}$ 為特解.

(合肥工業(yè)大學(xué),2020)設(shè) ${W=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,其中
${ \alpha_{1}=(1,2,-1,0,4)^{\prime},\alpha_{2}=(-1,3,2,4,1)^{\prime},\alpha_{3}=(2,9,-1,4,13)^{\prime} }$

求以 ${W}$ 作為其解空間的齊次線性方程組;

求以 ${W_{1}=\{\eta+\alpha | \alpha \in W\}}$ 為解集的非齊次線性方程組,其中 ${\eta=(1,2,1,2,1)^{\prime}}$ .

solution

首先記 ${A=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,則
${ A^{\prime}=\left(\begin{array}{c} \alpha_{1}^{\prime} \\ \alpha_{2}^{\prime} \\ \alpha_{3}^{\prime} \end{array}\right)=\left(\begin{array}{ccccc} 1 & 2 & -1 & 0 & 4 \\ -1 & 3 & 2 &4 & 1 \\ 2 & 9 & -1 & 4 & 13 \end{array}\right) . }$
對(duì) ${A^{\prime}}$ 進(jìn)行初等行變換化為階梯形有
${ A^{\prime} \longrightarrow\left(\begin{array}{ccccc} 1 & 2 & -1 & 0 & 4 \\ 0 &5 & 1 & 4 & 5 \\ 0 &0 &0 & 0 & 0 \end{array}\right) }$
因此 ${r\left(A^{\prime}\right)=2}$ ,結(jié)合 ${\alpha_{1},\alpha_{2}}$ 線性無(wú)關(guān)可知 ${\alpha_{1},\alpha_{2}}$ 是 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 的一個(gè)極大線性無(wú)關(guān)組,從而 ${W=L\left(\alpha_{1},\alpha_{2}\right)}$ .同時(shí)根據(jù)階梯形,可得方程組 ${A^{\prime} X=0}$ 的一組基礎(chǔ)解系為
${ \eta_{1}=(7,-1,5,0,0)^{\prime},\eta_{2}=(8,-4,0,5,0)^{\prime},\eta_{3}=(-2,-1,0,0,1)^{\prime} . }$
此時(shí)顯然有 ${A^{\prime} \eta_{i}=0(i=1,2,3)}$ ,記 ${B=\left(\eta_{1},\eta_{2},\eta_{3}\right)}$ ,即 ${A^{\prime} B=O}$ ,于是取轉(zhuǎn)置可知 ${B^{\prime} A=O}$ ,這說(shuō)明 ${A}$ 的列向量(即 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ ) 都是方程組 ${B^{\prime} X=0}$ 的解,而顯然 ${r\left(B^{\prime}\right)=3}$ ,從而 ${B^{\prime} X=0}$ 的基礎(chǔ)解系中含有 ${5-r\left(B^{\prime}\right)=2}$ 個(gè)向量,且 ${\alpha_{1},\alpha_{2}}$ 已經(jīng)是 ${B^{\prime} X=0}$ 的兩個(gè)線性無(wú)關(guān)的解向量,從而 ${\alpha_{1},\alpha_{2}}$ 就是 ${B^{\prime} X=0}$ 的基礎(chǔ)解系,這說(shuō)明 ${W}$ 就是 ${B^{\prime} X=0}$ 的解空間.故 ${B^{\prime} X=0}$ 即
${ \left\{\begin{array}{l} 7 x_{1}-x_{2}+5 x_{3}=0 \\ 8 x_{1}-4 x_{2}+5 x_{4}=0 \\ -2 x_{1}-x_{2}+x_{5}=0 \end{array}\right. }$
是滿足條件的一個(gè)方程組.
當(dāng) ${x_{1}=1,x_{2}=2,x_{3}=1,x_{4}=2,x_{5}=1}$ 時(shí),有
${ 7 x_{1}-x_{2}+5 x_{3}=10,8 x_{1}-4 x_{2}+5 x_{4}=10,-2 x_{1}-x_{2}+x_{5}=-3 }$
從而 ${\eta=(1,2,1,2,1)^{\prime}}$ 就是方程組
${ \left\{\begin{array}{l} 7 x_{1}-x_{2}+5 x_{3}=10 \\ 8 x_{1}-4 x_{2}+5 x_{4}=10 \\ -2 x_{1}-x_{2}+x_{5}=-3 . \end{array}\right. }$
的一個(gè)特解,而由第一問(wèn)可知此方程組導(dǎo)出組的解空間為 ${W}$ ,從而根據(jù)非齊次線性方程組解的性質(zhì)可知 ${W_{1}=\{\eta+\alpha | \alpha \in W\}}$ 就是上述方程組的解集.

求線性空間的和與交

(浙江工商大學(xué),2020)已知向量組
${ \begin{array}{c} \alpha_{1}=(1,2,-1,-2),\alpha_{2}=(3,1,1,1),\alpha_{3}=(-1,0,1,-1) \\ \beta_{1}=(2,5,-6,-5),\beta_{2}=(-1,2,-7,3) \end{array} }$
記子空間 ${V_{1}=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ 及 ${V_{2}=L\left(\beta_{1},\beta_{2}\right)}$ .求 ${V_{1}+V_{2}}$ 及 ${V_{1} \cap V_{2}}$ 的維數(shù)和基.

solution
首先記矩陣 ${A=\left(\alpha_{1}^{\prime},\alpha_{2}^{\prime},\alpha_{3}^{\prime},\beta_{1}^{\prime},\beta_{2}^{\prime}\right)}$ ,對(duì) ${A}$ 進(jìn)行初等行變換,化為階梯形,有
$\begin{aligned} A &=\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 2 & 1 & 0 & 5 & 2 \\ -1 & 1 & 1 & -6 & -7 \\ -2 & 1 & -1 & -5 & 3 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & -5 & 2 & 1 & 4 \\ 0 & 4 & 0 & -4 & -8 \\ 0 & 7 & -3 & -1 & 1 \end{array}\right) \\ & \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 2 & -4 & -6 \\ 0 & 0 & -3 & 6 & 15 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 0 & 0 & 6 \end{array}\right) \end{aligned}$
由此可知 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{2}}$ 為向量組 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2}}$ 的極大線性無(wú)關(guān)組,而明顯 ${V_{1}+V_{2}=L\left(\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2}\right)}$ ,所以 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{2}}$ 為 ${V_{1}+V_{2}}$ 的一組基,且 ${\operatorname{dim}\left(V_{1}+V_{2}\right)=3}$ .
另外,對(duì)任意的 ${\alpha \in V_{1} \cap V_{2}}$ ,不妨設(shè)
${ \alpha=k_{1} \alpha_{1}+k_{2} \alpha_{2}+k_{3} \alpha_{3}=l_{1} \beta_{1}+l_{2} \beta_{2} . }$
則有
${ k_{1} \alpha_{1}^{\prime}+k_{2} \alpha_{2}^{\prime}+k_{3} \alpha_{3}^{\prime}-l_{1} \beta_{1}^{\prime}-l_{2} \beta_{2}^{\prime}=0 . }$
將上式看作關(guān)于 ${k_{1},k_{2},k_{3},-l_{1},-l_{2}}$ 的線性方程組,則其系數(shù)矩陣為 ${A}$ ,而根據(jù)上述的階梯形可知方程組 ${A X=0}$ 的通解為
${ \left(k_{1},k_{2},k_{3},-l_{1},-l_{2}\right)^{\prime}=k(-3,1,2,1,0)^{\prime} . }$
其中 ${k}$ 為任意常數(shù),由此可知 ${l_{1}=-k,l_{2}=0}$ ,所以 ${\alpha=-k \beta_{1}}$ .這說(shuō)明 ${\beta_{1}}$ 即為 ${V_{1} \cap V_{2}}$ 的基,且 ${\operatorname{dim}\left(V_{1} \cap V_{2}\right)=1}$ .

(武漢理工大學(xué),2021)設(shè)向量組
${ \begin{array}{c} A: \alpha_{1}=(1,0,2),\alpha_{2}=(1,1,3),\alpha_{3}=(1,-1,a+2) ; \\ B: \beta_{1}=(1,2,a+3),\beta_{2}=(2,1,a+6),\beta_{3}=(2,1,a+4) . \end{array} }$

求 ${a}$ 滿足什么條件時(shí),向量組 ${A,B}$ 等價(jià);

求 ${a}$ 滿足什么條件時(shí),使得向量組 ${A,B}$ 不等價(jià);

記向量組 ${A,B}$ 生成的子空間分別為 ${W_{1},W_{2}}$ ,當(dāng) ${A,B}$ 不等價(jià)時(shí),求 ${W_{1}+W_{2}}$ 的基與維數(shù).

solution
首先記 ${P=\left(\alpha_{1}^{\prime},\alpha_{2}^{\prime},\alpha_{3}^{\prime}\right),Q=\left(\beta_{1}^{\prime},\beta_{2}^{\prime},\beta_{3}^{\prime}\right)}$ ,對(duì) ${P,Q}$ 進(jìn)行初等行變換,化為階梯形,有
$\begin{array}{c} P=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 2 & 3 & a+2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & a \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & a+1 \end{array}\right) \\ Q=\left(\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 1 \\ a+3 & a+6 & a+4 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & -a & -a-2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right) . \end{array}$
由此可知 ${\beta_{1},\beta_{2},\beta_{3}}$ 線性無(wú)關(guān).

當(dāng) ${a \neq-1}$ 時(shí),顯然 ${P,Q}$ 可逆,從而向量組 ${A,B}$ 等價(jià).
當(dāng) ${a=-1}$ 時(shí),由于 ${r(P)=2,r(Q)=3}$ ,所以此時(shí)向量組 ${A,B}$ 不等價(jià).
由 (2) 可知 ${a=-1}$ ,此時(shí)對(duì) ${(P,Q)}$ 進(jìn)行初等行變換,化為階梯形,有
$(P,Q)=\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 2 & 3 & 1 & 2 & 5 & 3 \end{array}\right) \rightarrow\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 0 & 1 & -1 & 0 & 1 & -1 \end{array}\right) \rightarrow\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 0 & 0 & 0 & -2 & 0 & -2 \end{array}\right)$
由此可知 ${\alpha_{1},\alpha_{2},\beta_{1}}$ 是 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2},\beta_{3}}$ 的極大線性無(wú)關(guān)組,進(jìn)而也是 ${V_{1}+V_{2}=L\left(\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2},\beta_{3}\right)}$ 的一組基,且 ${\operatorname{dim}\left(V_{1}+V_{2}\right)=3}$ .