【线性代数基础进阶】矩阵-part2

文章目录

三、初等变换、初等矩阵
- 矩阵的初等行变换
- 初等矩阵
- - 初等矩阵的逆
- 行阶梯矩阵
- 行最简矩阵
- 矩阵等价
四、分块矩阵
- 分四块
- 行、列分块
五、方阵的行列式

三、初等变换、初等矩阵

矩阵的初等行变换

用非000常数kkk乘AAA某行的每个元素，即倍乘
互换AAA中两行元素的位置，即互换
把AAA中某行所有元素的kkk倍加到另一行的对应元上，即倍加

初等矩阵

单位矩阵经过一次初等变换所得到的矩阵称为初等矩阵

初等矩阵PPP左乘矩阵AAA，其乘积PAPAPA就是矩阵AAA作一次与PPP同样的行变换
初等矩阵PPP右乘矩阵AAA，其乘积APAPAP就是矩阵AAA作一次与PPP同样的列变换

初等矩阵的逆

(100210001)(100−210001)=(100010001)\begin{pmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} ⎝⎛120010001⎠⎞⎝⎛1−20010001⎠⎞=⎝⎛100010001⎠⎞
初等矩阵若为倍加矩阵，则其逆矩阵为将其倍加的元改为其相反数，例如
(100210001)−1=(100−210001)\begin{pmatrix} 1 & 0 & 0 \\ \boxed{2} & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ \boxed{-2} & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} ⎝⎛120010001⎠⎞−1=⎝⎛1−20010001⎠⎞

(010100001)(010100001)=(100010001)\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} ⎝⎛010100001⎠⎞⎝⎛010100001⎠⎞=⎝⎛100010001⎠⎞
初等矩阵若为互换矩阵，则其逆矩阵为本身，例如
(010100001)−1=(010100001)\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} ⎝⎛010100001⎠⎞−1=⎝⎛010100001⎠⎞

(100050001)(1000150001)=(100010001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{5} & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} ⎝⎛100050001⎠⎞⎝⎛1000510001⎠⎞=⎝⎛100010001⎠⎞
初等矩阵若为倍乘矩阵，则其逆矩阵为倍乘的元的倒数，例如
(100050001)−1=(1000150001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & \boxed{5} & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \boxed{\frac{1}{5}} & 0 \\ 0 & 0 & 1 \end{pmatrix} ⎝⎛100050001⎠⎞−1=⎝⎛1000510001⎠⎞

初等矩阵均可逆，且其逆是同一类型的初等矩阵

例：已知aij≠0a_{ij}\ne0aij=0，如果(a11a12a13a21a22a23a31a32a33)P=(a112a12a12+a13a11a21a22a22+a23a11a31a32a32+a33)\begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}P=\begin{pmatrix}a_{11}^{2} & a_{12} & a_{12}+a_{13} \\ a_{11}a_{21} & a_{22} & a_{22}+a_{23} \\ a_{11}a_{31} & a_{32} & a_{32}+a_{33}\end{pmatrix}⎝⎛a11a21a31a12a22a32a13a23a33⎠⎞P=⎝⎛a112a11a21a11a31a12a22a32a12+a13a22+a23a32+a33⎠⎞，则P=()P=()P=()

经过第一列乘a11a_{11}a11倍，第三列加第二列，因此
P=(a1100011001)P=\begin{pmatrix} a_{11} & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} P=⎝⎛a1100010011⎠⎞

例：已知A=(aij)A=(a_{ij})A=(aij)是三阶矩阵，∣A∣=2|A|=2∣A∣=2，把矩阵AAA的第二行的−5-5−5倍加到第三行得到矩阵BBB，则(3BA∗)−1=()(3BA^{*})^{-1}=()(3BA∗)−1=()

由题意B=PAB=PAB=PA，且
P=(1000100−51)P=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -5 & 1 \end{pmatrix} P=⎝⎛10001−5001⎠⎞
有
3BA∗=3PAA∗=3P(∣A∣E)=6P3BA^{*}=3PAA^{*}=3P(|A|E)=6P 3BA∗=3PAA∗=3P(∣A∣E)=6P
则
(3BA∗)−1=(6P)−1=16P−1=16(100010051)(3BA^{*})^{-1}=(6P)^{-1}=\frac{1}{6}P^{-1}=\frac{1}{6}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 1 \end{pmatrix} (3BA∗)−1=(6P)−1=61P−1=61⎝⎛100015001⎠⎞

行阶梯矩阵

设AAA为m×nm\times nm×n矩阵，若满足

矩阵如有零行，则零行都在矩阵的底部
每个非零行的主元（即该行最左边的第111个非000元）所在列的下面的元素都是000

则称AAA为行阶梯矩阵

行最简矩阵

设AAA为m×nm\times nm×n矩阵，若AAA是行阶梯矩阵，且还满足

非零行的主元都是111，且主元所在列的其他元素都是000

则称AAA为行最简矩阵

矩阵等价

矩阵AAA经过有限次初等变换得到矩阵BBB就称矩阵AAA与矩阵BBB等价，记作A=∼BA\overset{\sim}{=}BA=∼B

矩阵A,BA,BA,B都是m×nm\times nm×n的矩阵，则A,BA,BA,B等价⇔r(A)=r(B)\Leftrightarrow r(A)=r(B)⇔r(A)=r(B)

设AAA是m×nm\times nm×n矩阵，则存在mmm姐可逆矩阵PPP和nnn阶可逆矩阵AAA，使
PAQ=(ErOOO)PAQ=\begin{pmatrix} E_{r} & O \\ O & O \end{pmatrix} PAQ=(ErOOO)

例：已知A=(1−1−132−1−3132−52)A=\begin{pmatrix}1 & -1 & -1 & 3 \\ 2 & -1 & -3 & 1 \\ 3 & 2 & -5 & 2\end{pmatrix}A=⎝⎛123−1−12−1−3−5312⎠⎞化其为行最简

A=(1−1−132−1−3132−52)→(1−1−1301−1−505−2−7)→(10−2−201−1−500318)从左到右各列上下同时开工→(10−2−201−1−50016)如果第一个非零元不是1，先化成1→(1001001010016)\begin{aligned} A&=\begin{pmatrix}1 & -1 & -1 & 3 \\ 2 & -1 & -3 & 1 \\ 3 & 2 & -5 & 2\end{pmatrix}\rightarrow \begin{pmatrix} 1 & -1 & -1 & 3 \\ 0 & 1 & -1 & -5 \\ 0 & 5 & -2 & -7 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1 & 0 & -2 & -2 \\ 0 & 1 & -1 & -5 \\ 0 & 0 & 3 & 18 \end{pmatrix}从左到右各列上下同时开工\\ &\rightarrow \begin{pmatrix} 1 & 0 & -2 & -2 \\ 0 & 1 & -1 & -5 \\ 0 & 0 & 1 & 6 \end{pmatrix}如果第一个非零元不是1，先化成1\\ &\rightarrow \begin{pmatrix} 1 & 0 & 0 & 10 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 6 \end{pmatrix} \end{aligned} A=⎝⎛123−1−12−1−3−5312⎠⎞→⎝⎛100−115−1−1−23−5−7⎠⎞→⎝⎛100010−2−13−2−518⎠⎞从左到右各列上下同时开工→⎝⎛100010−2−11−2−56⎠⎞如果第一个非零元不是1，先化成1→⎝⎛1000100011016⎠⎞

对于已知PA=B,A,BPA=B,A,BPA=B,A,B，求PPP的题
已知
A⟶行变换BA\overset{行变换}{\longrightarrow}B A⟶行变换B
即存在
Pτ⋯P2P1A=BP_{\tau}\cdots P_{2}P_{1}A=B Pτ⋯P2P1A=B
记P=Pτ⋯P2P1P=P_{\tau}\cdots P_{2}P_{1}P=Pτ⋯P2P1
Pτ⋯P2P1E=PP_{\tau}\cdots P_{2}P_{1}E=P Pτ⋯P2P1E=P
即A→BA\rightarrow BA→B同时E→PE\rightarrow PE→P
(A,E)⟶行变换(B,P)(A,E)\overset{行变换}{\longrightarrow}(B,P) (A,E)⟶行变换(B,P)

对于上一道例题，要求使得AAA变为行最简的PPP

(A∣E)=(1−1−1312−1−31132−521)→(1−1−13101−1−5−2105−2−7−31)→(1−1−13101−1−5−21003187−51)→(10−2−2−111−1−5−211673−5313)→(10010113−7323010113−2313001673−5313)\begin{aligned} (A|E)&=\begin{pmatrix} 1 & -1 & -1 & 3 & 1 & & \\ 2 & -1 & -3 & 1 & & 1 & \\ 3 & 2 & -5 & 2 & & & 1 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1 & -1 & -1 & 3 & 1 & & \\ 0 & 1 & -1 & -5 & -2 & 1 & \\ 0 & 5 & -2 & -7 & -3 & & 1 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1 & -1 & -1 & 3 & 1 & & \\ 0 & 1 & -1 & -5 & -2 & 1 & \\ 0 & 0 & 3 & 18 & 7 & -5 & 1 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1 & 0 & -2 & -2 & -1 & 1 & \\ & 1 & -1 & -5 & -2 & 1 & \\ & & 1 & 6 & \frac{7}{3} & - \frac{5}{3} & \frac{1}{3} \end{pmatrix}\\ & \rightarrow \begin{pmatrix} 1 & 0 & 0 & 10 & \frac{11}{3} & - \frac{7}{3} & \frac{2}{3} \\ 0 & 1 & 0 & 1 & \frac{1}{3} & - \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 1 & 6 & \frac{7}{3} & - \frac{5}{3} & \frac{1}{3} \end{pmatrix} \end{aligned} (A∣E)=⎝⎛123−1−12−1−3−5312111⎠⎞→⎝⎛100−115−1−1−23−5−71−2−311⎠⎞→⎝⎛100−110−1−133−5181−271−51⎠⎞→⎝⎛101−2−11−2−56−1−23711−3531⎠⎞→⎝⎛10001000110163113137−37−32−35323131⎠⎞

四、分块矩阵

分四块（AB,An,A−1AB,A^{n},A^{-1}AB,An,A−1），列分块、行分块（方程组的解，向量，阶）

分四块

对矩阵适当的分块处理，有以下的运算法则
(A1A2A3A4)+(B1B2B3B4)=(A1+B1A2+B2A3+B3A4+B4)(ABCD)(XYZW)=(AX+BZAY+BWCX+DZCY+DW)(ABCD)T=(ATCTBTDT)\begin{gathered} \begin{pmatrix} A_{1} & A_{2} \\ A_{3} & A_{4} \end{pmatrix}+\begin{pmatrix} B_{1} & B_{2} \\ B_{3} & B_{4} \end{pmatrix}=\begin{pmatrix} A_{1}+B_{1} & A_{2}+B_{2} \\ A_{3}+B_{3} & A_{4}+B_{4} \end{pmatrix}\\ \begin{pmatrix} A & B \\ C & D \end{pmatrix}\begin{pmatrix} X & Y \\ Z & W \end{pmatrix}=\begin{pmatrix} AX+BZ & AY+BW \\ CX+DZ & CY+DW \end{pmatrix}\\ \begin{pmatrix} A & B \\ C & D \end{pmatrix}^{T}=\begin{pmatrix} A^{T} & C^{T} \\ B^{T} & D^{T} \end{pmatrix} \end{gathered} (A1A3A2A4)+(B1B3B2B4)=(A1+B1A3+B3A2+B2A4+B4)(ACBD)(XZYW)=(AX+BZCX+DZAY+BWCY+DW)(ACBD)T=(ATBTCTDT)
如果B,CB,CB,C都是方阵
(BOOC)n=(BnOOCn)\begin{pmatrix} B & O \\ O & C \end{pmatrix}^{n}=\begin{pmatrix} B^{n} & O \\ O & C^{n} \end{pmatrix} (BOOC)n=(BnOOCn)
如果B,CB,CB,C是可逆方阵
(BOOC)−1=(B−1OOC−1),(OBCO)−1=(OC−1B−1O)\begin{pmatrix} B & O \\ O & C \end{pmatrix}^{-1}=\begin{pmatrix} B^{-1} & O \\ O & C^{-1} \end{pmatrix},\begin{pmatrix} O & B \\ C & O \end{pmatrix}^{-1}=\begin{pmatrix} O & C^{-1} \\ B^{-1} & O \end{pmatrix} (BOOC)−1=(B−1OOC−1),(OCBO)−1=(OB−1C−1O)

行、列分块

若AB=CAB=CAB=C
(γ1γ2γ3)(b11b12b13b21b22b23b31b32b33)=(δ1δ2δ3)\begin{pmatrix} \gamma_{1} & \gamma_{2} & \gamma_{3} \end{pmatrix}\begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix}=\begin{pmatrix} \delta_{1} & \delta_{2} & \delta_{3} \end{pmatrix} (γ1γ2γ3)⎝⎛b11b21b31b12b22b32b13b23b33⎠⎞=(δ1δ2δ3)
对应行列式
{b11γ1+b21γ2+b31γ3=δ1b12γ1+b22γ2+b32γ3=δ2b13γ1+b23γ2+b33γ3=δ3\begin{cases} b_{11}\gamma_{1}+b_{21}\gamma_{2}+b_{31}\gamma_{3}=\delta_{1} \\ b_{12}\gamma_{1}+b_{22}\gamma_{2}+b_{32}\gamma_{3}=\delta_{2} \\ b_{13}\gamma_{1}+b_{23}\gamma_{2}+b_{33}\gamma_{3}=\delta_{3} \\ \end{cases} ⎩⎨⎧b11γ1+b21γ2+b31γ3=δ1b12γ1+b22γ2+b32γ3=δ2b13γ1+b23γ2+b33γ3=δ3
即C=ABC=ABC=AB的列向量可由AAA的列向量线性表出
同理
(a11a12a13a21a22a23a31a32a33)(α1α2α3)=(β1β2β3)\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}\begin{pmatrix} \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{pmatrix}=\begin{pmatrix} \beta_{1} \\ \beta_{2} \\ \beta_{3} \end{pmatrix} ⎝⎛a11a21a31a12a22a32a13a23a33⎠⎞⎝⎛α1α2α3⎠⎞=⎝⎛β1β2β3⎠⎞
对应行列式
{a11α1+a12α2+a13α3=β1a21α1+a22α2+a23α3=β2a31α1+a32α2+a33α3=β3\begin{cases} a_{11}\alpha_{1}+a_{12}\alpha_{2}+a_{13}\alpha_{3}=\beta_{1} \\ a_{21}\alpha_{1}+a_{22}\alpha_{2}+a_{23}\alpha_{3}=\beta _{2} \\ a_{31}\alpha_{1}+a_{32}\alpha_{2}+a_{33}\alpha_{3}=\beta _{3} \\ \end{cases} ⎩⎨⎧a11α1+a12α2+a13α3=β1a21α1+a22α2+a23α3=β2a31α1+a32α2+a33α3=β3
即C=ABC=ABC=AB的行向量可由BBB的行向量线性表出

如果AAA可逆，AB=C→A−1C=BAB=C\rightarrow A^{-1}C=BAB=C→A−1C=B
BBB的行向量可由CCC的行向量线性表出
如果BBB可逆，AB=C→CB−1=AAB=C\rightarrow CB^{-1}=AAB=C→CB−1=A
AAA的列向量可由CCC的列向量线性表出

如果AB=CAB=CAB=C
A(β1β2β3)=(γ1γ2γ3)(Aβ1Aβ2Aβ3)=(γ1γ2γ3)\begin{aligned} A \begin{pmatrix} \beta_{1} & \beta_{2} & \beta_{3} \end{pmatrix}&=\begin{pmatrix} \gamma_{1} & \gamma_{2} & \gamma_{3} \end{pmatrix}\\ \begin{pmatrix} A \beta_{1} & A \beta_{2} & A \beta_{3} \end{pmatrix}&=\begin{pmatrix} \gamma_{1} & \gamma_{2} & \gamma_{3} \end{pmatrix} \end{aligned} A(β1β2β3)(Aβ1Aβ2Aβ3)=(γ1γ2γ3)=(γ1γ2γ3)
有
Aβ1=γ1,Aβ2=γ2,Aβ3=γ3A \beta_{1}=\gamma_{1},A \beta_{2}=\gamma_{2},A \beta_{3}=\gamma_{3} Aβ1=γ1,Aβ2=γ2,Aβ3=γ3
可知
β1\beta_{1}β1是方程组Ax=γ1Ax=\gamma_{1}Ax=γ1的解
β2\beta_{2}β2是方程组Ax=γ2Ax=\gamma_{2}Ax=γ2的解
β3\beta_{3}β3是方程组Ax=γ3Ax=\gamma_{3}Ax=γ3的解

如果AB=OAB=OAB=O
A(β1β2β3)=(OOOa)A \begin{pmatrix} \beta_{1} & \beta_{2} & \beta_{3} \end{pmatrix}=\begin{pmatrix} O & O & Oa \end{pmatrix} A(β1β2β3)=(OOOa)
可知β1,β2,β3\beta_{1},\beta_{2},\beta_{3}β1,β2,β3是Ax=OAx=OAx=O的解

例：已知X=ABAX=ABAX=ABA，其中A=(1001011001−10100−1),B=(0001001001001000)A=\begin{pmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1\end{pmatrix},B=\begin{pmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix}A=⎝⎛1001011001−10100−1⎠⎞,B=⎝⎛0001001001001000⎠⎞，则X=()X=()X=()

本题直接求也行
令
A=(1001011001−10100−1)=(ECC−E),B=(0001001001001000)=(OCCO)A=\left(\begin{array}{cc:cc} 1&0&0&1\\ 0&1&1&0\\\hdashline 0&1&-1&0\\ 1&0&0&-1 \end{array}\right)=\begin{pmatrix} E & C \\ C & -E \end{pmatrix},B=\left(\begin{array}{cc:cc}0&0&0&1\\ 0&0&1&0\\\hdashline 0&1&0&0\\ 1&0&0&0\end{array}\right)=\begin{pmatrix} O & C \\ C & O \end{pmatrix} A=⎝⎛1001011001−10100−1⎠⎞=(ECC−E),B=⎝⎛0001001001001000⎠⎞=(OCCO)
则
X=(ECC−E)(OCCO)(ECC−E)=(EC−CE)(ECC−E)=(2EOO−2E)\begin{aligned} X&=\begin{pmatrix} E & C \\ C & -E \end{pmatrix}\begin{pmatrix} O & C \\ C & O \end{pmatrix}\begin{pmatrix} E & C \\ C & -E \end{pmatrix}\\ &=\begin{pmatrix} E & C \\ -C & E \end{pmatrix}\begin{pmatrix} E & C \\ C & -E \end{pmatrix}\\ &=\begin{pmatrix} 2E & O \\ O & -2E \end{pmatrix} \end{aligned} X=(ECC−E)(OCCO)(ECC−E)=(E−CCE)(ECC−E)=(2EOO−2E)
可得X=(22−2−2)X=\begin{pmatrix}2 & & & \\ & 2 & & \\ & & -2 & \\ & & & -2\end{pmatrix}X=⎝⎛22−2−2⎠⎞

例：设A=(12−24t33−11)A=\begin{pmatrix}1 & 2 & -2 \\ 4 & t & 3 \\ 3 & -1 & 1\end{pmatrix}A=⎝⎛1432t−1−231⎠⎞，BBB为三阶非零矩阵，且AB=OAB=OAB=O，则t=()t=()t=()

由于AB=OAB=OAB=O
AB=A(β1β2β3)=(Aβ1Aβ2Aβ3)=(OOO)AB=A \begin{pmatrix} \beta_{1} & \beta_{2} & \beta_{3} \end{pmatrix}=\begin{pmatrix} A \beta_{1} & A \beta_{2} & A \beta_{3} \end{pmatrix}=\begin{pmatrix} O & O & O \end{pmatrix} AB=A(β1β2β3)=(Aβ1Aβ2Aβ3)=(OOO)
可知Aβ1=O,Aβ2=O,Aβ3=OA \beta_{1}=O,A \beta_{2}=O,A \beta_{3}=OAβ1=O,Aβ2=O,Aβ3=O
因此，β1,β2,β3\beta_{1},\beta_{2},\beta_{3}β1,β2,β3是AX=OAX=OAX=O的解
又因为B≠0B\ne0B=0，故AX=OAX=OAX=O存在非000解，根据克拉默法则
∣A∣=∣12−24t33−11∣=5(t+3)=0|A|=\begin{vmatrix} 1 & 2 & -2 \\ 4 & t & 3 \\ 3 & -1 & 1 \end{vmatrix}=5(t+3)=0 ∣A∣=∣∣1432t−1−231∣∣=5(t+3)=0
解得t=−3t=-3t=−3

例：计算(123456789)(20001−1101)\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}\begin{pmatrix}2 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 1\end{pmatrix}⎝⎛147258369⎠⎞⎝⎛2010100−11⎠⎞

简单的矩阵乘复杂的矩阵，可以考虑复杂的进行分块，复杂的在左面进行列分块，在右面进行行分块

原式=(α1α2α3)(20001−1101)=(2α1+α3α2−α2+α3)=(52114512381)\begin{aligned} 原式&=\begin{pmatrix} \alpha_{1} & \alpha_{2} & \alpha_{3} \end{pmatrix}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 2\alpha_{1}+\alpha_{3} & \alpha_{2} & -\alpha_{2}+\alpha_{3} \end{pmatrix}\\ &=\begin{pmatrix} 5 & 2 & 1 \\ 14 & 5 & 1 \\ 23 & 8 & 1 \end{pmatrix} \end{aligned} 原式=(α1α2α3)⎝⎛2010100−11⎠⎞=(2α1+α3α2−α2+α3)=⎝⎛51423258111⎠⎞

五、方阵的行列式

∣AT∣=∣A∣|A^{T}|=|A|∣AT∣=∣A∣
∣kA∣=kn∣A∣|kA|=k^{n}|A|∣kA∣=kn∣A∣
∣AB∣=∣A∣⋅∣B∣|AB|=|A|\cdot|B|∣AB∣=∣A∣⋅∣B∣
∣A2∣=∣A∣2|A^{2}|=|A|^{2}∣A2∣=∣A∣2
∣A∗∣=∣A∣n−1|A^{*}|=|A|^{n-1}∣A∗∣=∣A∣n−1
∣A−1∣=∣A∣−1=1∣A∣|A^{-1}|=|A|^{-1}=\frac{1}{|A|}∣A−1∣=∣A∣−1=∣A∣1
∣AO∗B∣=∣A∗OB∣=∣A∣⋅∣B∣\begin{vmatrix}A&O\\ *&B\end{vmatrix}=\begin{vmatrix}A&*\\O&B\end{vmatrix}=|A|\cdot|B|∣∣A∗OB∣∣=∣∣AO∗B∣∣=∣A∣⋅∣B∣
∣OAB∗∣=∣∗ABO∣=(−1)mn∣A∣⋅∣B∣\begin{vmatrix}O&A\\B&*\end{vmatrix}=\begin{vmatrix}*&A\\B&O\end{vmatrix}=(-1)^{mn}|A|\cdot|B|∣∣OBA∗∣∣=∣∣∗BAO∣∣=(−1)mn∣A∣⋅∣B∣
如果A∼BA\sim BA∼B，则∣A∣=∣B∣,∣A+kE∣=∣B+kE∣|A|=|B|,|A+kE|=|B+kE|∣A∣=∣B∣,∣A+kE∣=∣B+kE∣，一般∣A+B∣≠∣A∣+∣B∣|A+B|\ne|A|+|B|∣A+B∣=∣A∣+∣B∣

例：A,BA,BA,B是nnn阶矩阵，∣A∣=2,∣B∣=−3|A|=2,|B|=-3∣A∣=2,∣B∣=−3，则

∣A∗B−1−A−1B∗∣=()|A^{*}B^{-1}-A^{-1}B^{*}|=()∣A∗B−1−A−1B∗∣=()
∣A∗B−1−A−1B∗∣=∣∣A∣A−1B−1−∣B∣A−1B−1∣=∣5A−1B−1∣=−5n6\begin{aligned} |A^{*}B^{-1}-A^{-1}B^{*}|&=||A|A^{-1}B^{-1}-|B|A^{-1}B^{-1}|\\ &=|5A^{-1}B^{-1}|\\ &=-\frac{5^{n}}{6} \end{aligned} ∣A∗B−1−A−1B∗∣=∣∣A∣A−1B−1−∣B∣A−1B−1∣=∣5A−1B−1∣=−65n

∣OATB∗2B∣=()\begin{vmatrix}O&A^{T}\\B^{*}&2B\end{vmatrix}=()∣∣OB∗AT2B∣∣=()
∣OATB∗2B∣=(−1)n⋅n∣AT∣∣B∗∣=(−1)n2+n−1⋅2⋅3n−1=−2⋅3n−1\begin{aligned} \begin{vmatrix}O&A^{T}\\B^{*}&2B\end{vmatrix}&=(-1)^{n\cdot n}|A^{T}||B^{*}|\\ &=(-1)^{n^{2}+n-1}\cdot 2\cdot 3^{n-1}\\ &=-2\cdot3^{n-1} \end{aligned} ∣∣OB∗AT2B∣∣=(−1)n⋅n∣AT∣∣B∗∣=(−1)n2+n−1⋅2⋅3n−1=−2⋅3n−1