【660-线性代数】补充秩为1的方阵性质 & 矩阵练习

287矩阵乘法

已知 A = ( 1 2 3 4 5 6 7 8 9 ) , Λ = ( 1 2 − 1 ) A=

,\Lambda= A=⎝ ⎛​147​258​369​⎠ ⎞​,Λ=⎝ ⎛​1​2​−1​⎠ ⎞​，则$A\Lambda -\Lambda A=()$

本题属于基本计算，不再展示过程

( 0 2 − 6 − 4 0 − 18 14 24 0 )

⎝ ⎛​0−414​2024​−6−180​⎠ ⎞​

可以用最基本的方法，不难
如果两个行列式相乘，其中一个比较简单，可以考虑矩阵的初等变换，例如本题
如果两个都复杂，可以考虑矩阵分块，即 ( 1 2 3 4 5 6 7 8 9 ) ( 2 0 0 0 1 − 1 1 0 1 ) = ( α 1 α 2 α 3 ) ( 2 0 0 0 1 − 1 1 0 1 ) = ( 2 α 1 + α 3 α 2 − α 2 + α 3 )

(123456789)(20001−1101)=(α1α2α3)(20001−1101)=(2α1+α3α2−α2+α3)" role="presentation" style="position: relative;">⎛⎝⎜147258369⎞⎠⎟⎛⎝⎜2010100−11⎞⎠⎟=(α1α2α3)⎛⎝⎜2010100−11⎞⎠⎟=(2α1+α3α2−α2+α3)$$\begin{array}{rl}\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\left(\begin{array}{ccc}2& 0& 0\\ 0& 1& -1\\ 1& 0& 1\end{array}\right)& =\left(\begin{array}{ccc}{\alpha }_1& {\alpha }_2& {\alpha }_3\end{array}\right)\left(\begin{array}{ccc}2& 0& 0\\ 0& 1& -1\\ 1& 0& 1\end{array}\right)\\ & =\left(\begin{array}{ccc}2{\alpha }_1+{\alpha }_3& {\alpha }_2& -{\alpha }_2+{\alpha }_3\end{array}\right)\end{array}$$

⎝ ⎛​147​258​369​⎠ ⎞​⎝ ⎛​201​010​0−11​⎠ ⎞​​=(α1​​α2​​α3​​)⎝ ⎛​201​010​0−11​⎠ ⎞​=(2α1​+α3​​α2​​−α2​+α3​​)​

289秩为$1$的矩阵乘法

已知 A = ( 2 − 1 3 4 − 2 6 − 2 1 − 3 ) A=

A=⎝ ⎛​24−2​−1−21​36−3​⎠ ⎞​，则$A^{10}=()$

若$\alpha =(a_1,a_2,a_3{)}^T,\beta =(b_1,b_2,b_3{)}^T$，设
A = α β T = ( a 1 a 2 a 3 ) ( b 1 b 2 b 3 ) = ( a 1 b 1 a 1 b 2 a 1 b 3 a 2 b 1 a 2 b 2 a 2 b 3 a 3 b 1 a 3 b 2 a 3 b 3 ) B = β T α = ( b 1 b 2 b 3 ) ( a 1 a 2 a 3 ) = a 1 b 1 + a 2 b 2 + a 3 b 3

A=αβT=(a1a2a3)(b1b2b3)=(a1b1a1b2a1b3a2b1a2b2a2b3a3b1a3b2a3b3)B=βTα=(b1b2b3)(a1a2a3)=a1b1+a2b2+a3b3" role="presentation" style="position: relative;">AB=αβT=⎛⎝⎜a1a2a3⎞⎠⎟(b1b2b3)=⎛⎝⎜a1b1a2b1a3b1a1b2a2b2a3b2a1b3a2b3a3b3⎞⎠⎟=βTα=(b1b2b3)⎛⎝⎜a1a2a3⎞⎠⎟=a1b1+a2b2+a3b3$$\begin{array}{rl}A& =\alpha {\beta }^T=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\left(\begin{array}{ccc}{b}_1& b_2& b_3\end{array}\right)=\left(\begin{array}{ccc}{a}_1{b}_1& a_1{b}_2& a_1{b}_3\\ a_2{b}_1& a_2{b}_2& a_2{b}_3\\ a_3{b}_1& a_3{b}_2& a_3{b}_3\end{array}\right)\\ B& ={\beta }^T\alpha =\left(\begin{array}{ccc}{b}_1& b_2& b_3\end{array}\right)\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)=a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$$

AB​=αβT=⎝ ⎛​a1​a2​a3​​⎠ ⎞​(b1​​b2​​b3​​)=⎝ ⎛​a1​b1​a2​b1​a3​b1​​a1​b2​a2​b2​a3​b2​​a1​b3​a2​b3​a3​b3​​⎠ ⎞​=βTα=(b1​​b2​​b3​​)⎝ ⎛​a1​a2​a3​​⎠ ⎞​=a1​b1​+a2​b2​+a3​b3​​
$A$为秩为$1$的三阶矩阵，$B$为一个数
因此有，对于任意方阵$A$，如果$r(A)=1$，则有
A = α β T A 2 = α ( β T α ) β T = α l β T = l α β T = l A

A=αβTA2=α(βTα)βT=αlβT=lαβT=lA" role="presentation" style="position: relative;">AA2=αβT=α(βTα)βT=αlβT=lαβT=lA$$\begin{array}{rl}A& =\alpha {\beta }^T\\ A^2& =\alpha ({\beta }^T\alpha ){\beta }^T=\alpha l{\beta }^T=l\alpha {\beta }^T=lA\end{array}$$

AA2​=αβT=α(βTα)βT=αlβT=lαβT=lA​
其中$l={\beta }^T\alpha ={\alpha }^T\beta =\sum a_{ii}$，进而$A^m=l^{m-1}A$

观察本题，$r(A)=1$，因此有
$$A^2=lA$$
又因为$l=\sum a_{ii}=-3$，因此
$$A^{10}=l^{9}A=-3^{9}A$$

292伴随矩阵

若 A = ( 1 2 0 3 4 0 0 0 5 ) A=

A=⎝ ⎛​130​240​005​⎠ ⎞​，则矩阵$A$的伴随矩阵$A^{\ast }=()$

可以用最基本的方法
如果想用分块矩阵的方法要注意，
( A O O B ) − 1 = ( A − 1 O O B − 1 ) , ( A O O B ) ∗ ≠ ( A ∗ O O B ∗ )

(AOOB)" role="presentation" style="position: relative;">(AOOB)$$\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)$$

^{-1}=

(A−1OOB−1)" role="presentation" style="position: relative;">(A−1OOB−1)$$\left(\begin{array}{cc}{A}^{-1}& O\\ O& B^{-1}\end{array}\right)$$

,

(AOOB)" role="presentation" style="position: relative;">(AOOB)$$\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)$$

^{*}\ne

(A∗OOB∗)" role="presentation" style="position: relative;">(A∗OOB∗)$$\left(\begin{array}{cc}{A}^{\ast }& O\\ O& B^{\ast }\end{array}\right)$$

(AO​OB​)−1=(A−1O​OB−1​),(AO​OB​)∗=(A∗O​OB∗​)
因此想要用公式求伴随矩阵，要用
$$A{A}^{\ast }=|A|E\Rightarrow A^{\ast }=|A|A^{-1}$$
变为逆矩阵才能用分块矩阵公式

A ∗ = ∣ 1 2 0 3 4 0 0 0 5 ∣ ( − 2 1 0 3 2 − 1 2 0 0 0 1 5 ) = − 10 ( − 2 1 0 3 2 − 1 2 0 0 0 1 5 ) = ( 20 − 10 0 − 15 5 0 0 0 − 2 )

A∗​=∣ ∣​130​240​005​∣ ∣​⎝ ⎛​−223​0​1−21​0​0051​​⎠ ⎞​=−10⎝ ⎛​−223​0​1−21​0​0051​​⎠ ⎞​=⎝ ⎛​20−150​−1050​00−2​⎠ ⎞​​

293伴随矩阵、逆矩阵

设矩阵$A$的伴随矩阵 A ∗ = ( 4 − 2 0 0 − 3 1 0 0 0 0 − 4 0 0 0 0 − 1 ) A^{*}=

A∗=⎝ ⎛​4−300​−2100​00−40​000−1​⎠ ⎞​，则$A=()$

由于$A{A}^{\ast }=|A|E$，故
$$A=|A|(A^{\ast }{)}^{-1}$$

根据$A{A}^{\ast }=|A|E$，移项得$A^{-1}=\frac{1}{|A|}{A}^{\ast }$，显然$A,A^{\ast },A^{-1}$之间可以互相求

由已知得$|A^{\ast }|=-8$，又有$|A^{\ast }|=|A{|}^3$，得
$$|A|=\sqrt[3]{-8}=-2$$
又
$$
(A^{*}){-1}=

^{-1}=\begin{pmatrix}

• \frac{1}{2} & -1 & 0 & 0 \ - \frac{3}{2} & -2 & 0 & 0 \ 0 & 0 & - \frac{1}{4} & 0 \ 0 & 0 & 0 & -1
  \end{pmatrix}
  $$

求矩阵的逆矩阵，先考虑能不能用分块矩阵，或者二阶伴随矩阵口诀的方法

因此
A = ∣ A ∣ ( A ∗ ) − 1 = ( 1 2 0 0 3 4 0 0 0 0 1 2 0 0 0 0 2 ) A=|A|(A^{*})^{-1}=

A=∣A∣(A∗)−1=⎝ ⎛​1300​2400​0021​0​0002​⎠ ⎞​

297逆矩阵

已知$AB=A+B$，其中 B = ( 1 1 0 1 1 0 0 0 2 ) B=

B=⎝ ⎛​110​110​002​⎠ ⎞​，则$(A-E{)}^{-1}=()$

这里只讨论给出矩阵关系，未完全给出各个矩阵的数字表示的题目，一般的思路是，以本题为例，将题目的条件化简（别着急往里带），要求谁，就写成，
$$(A-E)(矩阵)=E$$
第一个括号内为所求逆矩阵对应的原矩阵，第二个括号内是为了和题目条件凑相等而得到的，主要是利用的观察法，等号另一侧只能有$E$或者其他给出了数字表示的矩阵（例如本题，还可以有$B$）

由$AB=A+B$，可得
$$(A-E)(B-E)=E$$
因此 ( A − E ) − 1 = B − E = ( 0 1 0 1 0 0 0 0 1 ) (A-E)^{-1}=B-E=

(A−E)−1=B−E=⎝ ⎛​010​100​001​⎠ ⎞​

298线性方程组、逆矩阵

已知${\alpha }_1=(1,0,0{)}^T,{\alpha }_2=(1,2,-1{)}^T,{\alpha }_3=(-1,1,0{)}^T$且$A{\alpha }_1=(2,1{)}^T,A{\alpha }_2=(-1,1{)}^T,A{\alpha }_3=(3,-4{)}^T$，则$A=()$

利用分块矩阵有
A ( α 1 , α 2 , α 3 ) = ( A α 1 , A α 2 , A α 3 ) = ( 2 − 1 3 1 1 − 4 ) A(\alpha_{1},\alpha_{2},\alpha_{3})=(A \alpha_{1},A \alpha_{2},A \alpha_{3})=

A(α1​,α2​,α3​)=(Aα1​,Aα2​,Aα3​)=(21​−11​3−4​)
其中
∣ α 1 , α 2 , α 3 ∣ = ∣ 1 1 − 1 0 2 1 0 − 1 0 ∣ = 1 ≠ 0 \left|\alpha_{1},\alpha_{2},\alpha_{3}\right|=

|11−10210−10|" role="presentation" style="position: relative;">∣∣∣∣10012−1−110∣∣∣∣$$\left|\begin{array}{ccc}1& 1& -1\\ 0& 2& 1\\ 0& -1& 0\end{array}\right|$$

=1\ne 0 ∣α1​,α2​,α3​∣=∣ ∣​100​12−1​−110​∣ ∣​=1=0
因此$({\alpha }_1,{\alpha }_2,{\alpha }_3)$可逆
A = ( 2 − 1 3 1 1 − 4 ) ⋅ ( 1 1 − 1 0 2 1 0 − 1 0 ) − 1 = ( 2 − 1 3 1 1 − 4 ) ⋅ ( 1 1 3 0 0 − 1 0 1 2 ) = ( 2 5 13 1 − 3 − 6 )

A=(2−1311−4)⋅(11−10210−10)−1=(2−1311−4)⋅(11300−1012)=(25131−3−6)" role="presentation" style="position: relative;">A=(21−113−4)⋅⎛⎝⎜10012−1−110⎞⎠⎟−1=(21−113−4)⋅⎛⎝⎜1001013−12⎞⎠⎟=(215−313−6)$$\begin{array}{rl}A& =\left(\begin{array}{ccc}2& -1& 3\\ 1& 1& -4\end{array}\right)\cdot {\left(\begin{array}{ccc}1& 1& -1\\ 0& 2& 1\\ 0& -1& 0\end{array}\right)}^{-1}\\ & =\left(\begin{array}{ccc}2& -1& 3\\ 1& 1& -4\end{array}\right)\cdot \left(\begin{array}{ccc}1& 1& 3\\ 0& 0& -1\\ 0& 1& 2\end{array}\right)\\ & =\left(\begin{array}{ccc}2& 5& 13\\ 1& -3& -6\end{array}\right)\end{array}$$

A​=(21​−11​3−4​)⋅⎝ ⎛​100​12−1​−110​⎠ ⎞​−1=(21​−11​3−4​)⋅⎝ ⎛​100​101​3−12​⎠ ⎞​=(21​5−3​13−6​)​

如果矩阵不可逆，就涉及到线性方程组的方法

设
A = ( x 1 x 2 x 3 y 1 y 2 y 3 ) A=

A=(x1​y1​​x2​y2​​x3​y3​​)
又因为
A ( α 1 , α 2 , α 3 ) = ( 2 − 1 3 1 1 − 4 )

A(α1,α2,α3)=(2−1311−4)" role="presentation" style="position: relative;">A(α1,α2,α3)=(21−113−4)$$\begin{array}{r}A({\alpha }_1,{\alpha }_2,{\alpha }_3)=\left(\begin{array}{ccc}2& -1& 3\\ 1& 1& -4\end{array}\right)\end{array}$$

A(α1​,α2​,α3​)=(21​−11​3−4​)​
有
{ x 1 = 2 x 1 + 2 x 2 − x 3 = − 1 − x 1 + x 2 = 3 y 1 = 1 y 1 + 2 y 2 − y 3 = 1 − y 1 + y 2 = 1 \left\{

x1=2x1+2x2−x3=−1−x1+x2=3y1=1y1+2y2−y3=1−y1+y2=1" role="presentation" style="position: relative;">x1=2x1+2x2−x3=−1−x1+x2=3y1=1y1+2y2−y3=1−y1+y2=1$$\begin{array}{rl}& x_1=2\\ & x_1+2x_2-x_3=-1\\ & -x_1+x_2=3\\ & y_1=1\\ & y_1+2y_2-y_3=1\\ & -y_1+y_2=1\end{array}$$

\right. ⎩ ⎨ ⎧​​x1​=2x1​+2x2​−x3​=−1−x1​+x2​=3y1​=1y1​+2y2​−y3​=1−y1​+y2​=1​
对应增广矩阵
( 1 0 0 2 1 1 2 − 1 − 1 1 − 1 1 0 3 1 ) → ( 1 0 0 2 1 0 1 0 5 − 1 0 0 1 13 − 6 )

(1002112−1−11−11031)" role="presentation" style="position: relative;">⎛⎝⎜11−10210−102−13111⎞⎠⎟$$\left(\begin{array}{ccccc}1& 0& 0& 2& 1\\ 1& 2& -1& -1& 1\\ -1& 1& 0& 3& 1\end{array}\right)$$

\rightarrow

(100210105−100113−6)" role="presentation" style="position: relative;">⎛⎝⎜10001000125131−1−6⎞⎠⎟$$\left(\begin{array}{ccccc}1& 0& 0& 2& 1\\ 0& 1& 0& 5& -1\\ 0& 0& 1& 13& -6\end{array}\right)$$

⎝ ⎛​11−1​021​0−10​2−13​111​⎠ ⎞​→⎝ ⎛​100​010​001​2513​1−1−6​⎠ ⎞​
因此矩阵$A$为
( 2 5 13 1 − 3 − 6 )

(25131−3−6)" role="presentation" style="position: relative;">(215−313−6)$$\left(\begin{array}{ccc}2& 5& 13\\ 1& -3& -6\end{array}\right)$$

(21​5−3​13−6​)

如果系数矩阵是不可逆矩阵，这里是求不出详细的矩阵的

299线性方程组、逆矩阵

若 ( 1 1 2 2 ) A = ( 2 3 4 6 )

A= (12​12​)A=(24​36​)，则$A=()$

显然矩阵 ( 1 1 2 2 )

(12​12​)不可逆，故设 A = ( x 1 y 1 x 2 y 2 ) A=

(x1y1x2y2)" role="presentation" style="position: relative;">(x1x2y1y2)$$\left(\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right)$$

A=(x1​x2​​y1​y2​​)，有
( 1 1 2 2 ) ( x 1 y 1 x 2 y 2 ) = ( 2 3 4 6 )

(1122)" role="presentation" style="position: relative;">(1212)$$\left(\begin{array}{cc}1& 1\\ 2& 2\end{array}\right)$$

(x1y1x2y2)" role="presentation" style="position: relative;">(x1x2y1y2)$$\left(\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right)$$

=

(2346)" role="presentation" style="position: relative;">(2436)$$\left(\begin{array}{cc}2& 3\\ 4& 6\end{array}\right)$$

(12​12​)(x1​x2​​y1​y2​​)=(24​36​)
对应方程组
{ x 1 + x 2 = 2 y 1 + y 2 = 3 ⇒ { x 1 = 2 − t x 2 = t y 1 = 3 − u y 2 = u \left\{

x1+x2=2y1+y2=3" role="presentation" style="position: relative;">x1+x2=2y1+y2=3$$\begin{array}{rl}& x_1+x_2=2\\ & y_1+y_2=3\end{array}$$

\right.\Rightarrow \left\{

x1=2−tx2=ty1=3−uy2=u" role="presentation" style="position: relative;">x1=2−tx2=ty1=3−uy2=u$$\begin{array}{rl}& x_1=2-t\\ & x_2=t\\ & y_1=3-u\\ & y_2=u\end{array}$$

\right. {​x1​+x2​=2y1​+y2​=3​⇒⎩ ⎨ ⎧​​x1​=2−tx2​=ty1​=3−uy2​=u​

这里用的就是大的增广矩阵

所以
A = ( 2 − t 3 − u t u ) , t , u 是任意常数 A=

,t,u是任意常数 A=(2−tt​3−uu​),t,u是任意常数