快速求三階矩陣的逆矩陣
前言
一般情況下，我們求解伴隨矩陣是要注意符號(hào)問題和位置問題的（如下所示）
A
?
1
=
1
[
??
]
[
?
[
??
]
?
[
??
]
?
[
??
]
??
?
[
??
]
]
=
A
?
1
=
1
[
??
]
[
???
M
11
?
[
M
12
]
???
M
13
?
[
M
21
]
???
M
22
?
[
M
23
]
??
???
M
31
?
[
M
32
]
???
M
33
]
?
\begin{aligned} & A^{-1}=\frac{1}{[\ \ ]} \left[\begin{array}{cccccc} & -[\ \ ] & \\ -[\ \ ] & & -[\ \ ]\ \ \\ & -[\ \ ] & \\ \end{array}\right]= \\ \\ & A^{-1}=\frac{1}{[\ \ ]} \left[\begin{array}{cccccc} \ \ \ M_{11} & -[M_{12}] & \ \ \ M_{13}\\ -[M_{21}] & \ \ \ M_{22} & -[M_{23}]\ \ \\ \ \ \ M_{31} & -[M_{32}] & \ \ \ M_{33}\\ \end{array}\right]^\top\\ \end{aligned}
?A?1=[??]1?????[??]??[??]?[??]??[??]??????=A?1=[??]1???????M11??[M21?]???M31???[M12?]???M22??[M32?]????M13??[M23?]?????M33???????
我們根據(jù)位置安排(行調(diào)換)的策略可以避免符號(hào)問題，將問題進(jìn)行化簡(jiǎn)。
例題一
求矩陣
D
D
D 的逆矩陣
D
=
[
2
1
1
1
2
1
2
3
1
]
D=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 2 & 3 & 1 \end{array}\right]
D=???212?123?111????
我們把第一二列抄寫到矩陣后面
D
1
=
[
2
1
1
2
1
1
2
1
1
2
2
3
1
2
3
]
D_1=\left[\begin{array}{lll|ll} 2 & 1 & 1 & 2 & 1 \\ 1 & 2 & 1 & 1 & 2 \\ 2 & 3 & 1 & 2 & 3 \end{array}\right]
D1?=???212?123?111?212?123????
然后把第一二行抄寫到矩陣下面（新矩陣
D
1
D_1
D1? 的第一二行），
這樣我們就得到了一個(gè)五階矩陣：
D
2
=
[
2
1
1
2
1
1
2
1
1
2
2
3
1
2
3
2
1
1
2
1
1
2
1
1
2
]
=
[
2
1
1
2
1
1
2
1
1
2
2
3
1
2
3
2
1
1
2
1
1
2
1
1
2
]
D_2=\left[\begin{array}{lll|ll} 2 & 1 & 1 & 2 & 1 \\ 1 & 2 & 1 & 1 & 2 \\ 2 & 3 & 1 & 2 & 3 \\ \hline 2 & 1 & 1 & 2 & 1 \\ 1 & 2 & 1 & 1 & 2 \\ \end{array}\right]= \left[\begin{array}{lllll} \textcolor{cornflowerblue}{2} & \textcolor{cornflowerblue}{1} & \textcolor{cornflowerblue}{1} & \textcolor{cornflowerblue}{2} & \textcolor{cornflowerblue}{1} \\ \textcolor{cornflowerblue}{1} & 2 & 1 & 1 & 2 \\ \textcolor{cornflowerblue}{2} & 3 & 1 & 2 & 3 \\ \textcolor{cornflowerblue}{2} & 1 & 1 & 2 & 1 \\ \textcolor{cornflowerblue}{1} & 2 & 1 & 1 & 2 \\ \end{array}\right]
D2?=???????21221?12312?11111?21221?12312?????????=???????21221?12312?11111?21221?12312????????
然后我們把第一行和第一列刪除（新矩陣
D
2
D_2
D2? 的第一行和第一列，將標(biāo)藍(lán)的元素刪除）
D
3
=
[
2
1
1
2
3
1
2
3
1
1
2
1
2
1
1
2
]
D_3=\left[\begin{array}{lllll} 2 & 1 & 1 & 2 \\ 3 & 1 & 2 & 3 \\ 1 & 1 & 2 & 1 \\ 2 & 1 & 1 & 2 \\ \end{array}\right]
D3?=?????2312?1111?1221?2312??????
然后我們算出矩陣分塊組成的九個(gè)二階行列式：
然后我們將求出的九個(gè)行列式結(jié)果填充到伴隨矩陣的框架里，記得加上轉(zhuǎn)置符號(hào)
[
?
1
???
1
?
1
???
2
???
0
?
4
?
1
?
1
???
2
]
?
=
[
?
1
???
2
?
1
???
1
???
0
?
1
?
1
?
4
???
2
]
\left[\begin{array}{lllll} -1 &\ \ \ 1 & -1 \\ \ \ \ 2 &\ \ \ 0 & -4 \\ -1 & -1 &\ \ \ 2 \\ \end{array}\right]^\top= \left[\begin{array}{lllll} -1 &\ \ \ 2 & -1 \\ \ \ \ 1 &\ \ \ 0 & -1 \\ -1 & -4 &\ \ \ 2 \\ \end{array}\right]
????1???2?1????1???0?1??1?4???2?????=????1???1?1????2???0?4??1?1???2????
這樣我們就得到了伴隨矩陣，然后計(jì)算矩陣對(duì)應(yīng)的行列式
∣
D
∣
=
?
2
|D|=-2
∣D∣=?2，
最后根據(jù)公式
A
?
1
=
1
∣
A
∣
A
?
A^{-1}=\frac{1}{|A|}A^*
A?1=∣A∣1?A?，求出逆矩陣
D
?
1
D^{-1}
D?1
D
?
1
=
1
∣
D
∣
D
?
=
???
1
?
2
[
?
1
???
2
?
1
???
1
???
0
?
1
?
1
?
4
???
2
]
=
[
???
1
2
?
1
???
1
2
?
1
2
???
0
???
1
2
???
1
2
???
2
?
1
]
D^{-1}=\frac{1}{|D|}D^*=\frac{\ \ \ 1}{-2} \left[\begin{array}{lllll} -1 &\ \ \ 2 & -1 \\ \ \ \ 1 &\ \ \ 0 & -1 \\ -1 & -4 &\ \ \ 2 \\ \end{array}\right]= \left[\begin{array}{lllll} \ \ \ \frac{1}{2} &-1 & \ \ \ \frac{1}{2} \\ -\frac{1}{2} &\ \ \ 0 & \ \ \ \frac{1}{2} \\ \ \ \ \frac{1}{2} &\ \ \ 2 & -1 \\ \end{array}\right]
D?1=∣D∣1?D?=?2???1?????1???1?1????2???0?4??1?1???2????=??????21??21????21???1???0???2????21????21??1????文章來源：http://www.zghlxwxcb.cn/news/detail-406350.html
例題二
求矩陣
A
A
A 的逆矩陣
A
=
[
1
1
1
4
2
1
9
3
1
]
A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{array}\right]
A=???149?123?111????
抄寫后對(duì)應(yīng)的五階矩陣為：
A
1
=
[
1
1
1
1
1
4
2
1
4
2
9
3
1
9
3
1
1
1
1
1
4
2
1
4
2
]
A_1=\left[\begin{array}{lll} 1 & 1 & 1 &1 & 1\\ 4 & 2 & 1 &4 & 2\\ 9 & 3 & 1 &9 & 3\\ 1 & 1 & 1 &1 & 1\\ 4 & 2 & 1 &4 & 2 \end{array}\right]
A1?=???????14914?12312?11111?14914?12312????????
刪除后得到的四階矩陣為：
A
2
=
[
2
1
4
2
3
1
9
3
1
1
1
1
2
1
4
2
]
A_2=\left[\begin{array}{lll} 2 & 1 &4 & 2\\ 3 & 1 &9 & 3\\ 1 & 1 &1 & 1\\ 2 & 1 &4 & 2 \end{array}\right]
A2?=?????2312?1111?4914?2312??????
那么對(duì)應(yīng)的伴隨矩陣為：
A
?
=
[
?
1
???
5
?
6
???
2
?
8
???
6
?
1
???
3
?
2
]
?
=
[
?
1
???
2
?
1
???
5
?
8
???
3
?
6
???
6
?
2
]
A^*=\left[\begin{array}{lllll} -1 &\ \ \ 5 & -6 \\ \ \ \ 2 & -8 & \ \ \ 6 \\ -1 & \ \ \ 3 & -2 \\ \end{array}\right]^\top= \left[\begin{array}{lllll} -1 &\ \ \ 2 & -1 \\ \ \ \ 5 & -8 & \ \ \ 3 \\ -6& \ \ \ 6 & -2 \\ \end{array}\right]
A?=????1???2?1????5?8???3??6???6?2?????=????1???5?6????2?8???6??1???3?2????
矩陣對(duì)應(yīng)的行列式為
∣
A
∣
=
?
2
|A|=-2
∣A∣=?2，根據(jù)公式計(jì)算得到逆矩陣：
A
?
1
=
1
∣
A
∣
A
?
=
???
1
?
2
[
?
1
???
2
?
1
???
5
?
8
???
3
?
6
???
6
?
2
]
=
[
???
1
2
?
1
???
1
2
?
5
2
???
4
?
3
2
???
3
?
3
???
1
2
]
A^{-1}=\frac{1}{|A|}A^*=\frac{\ \ \ 1}{-2} \left[\begin{array}{lllll} -1 &\ \ \ 2 & -1 \\ \ \ \ 5 & -8 & \ \ \ 3 \\ -6& \ \ \ 6 & -2 \\ \end{array}\right]= \left[\begin{array}{lllll} \ \ \ \frac{1}{2} &-1 & \ \ \ \frac{1}{2} \\ -\frac{5}{2} &\ \ \ 4 & -\frac{3}{2} \\ \ \ \ 3 & -3 & \ \ \ \frac{1}{2} \\ \end{array}\right]
A?1=∣A∣1?A?=?2???1?????1???5?6????2?8???6??1???3?2????=??????21??25????3??1???4?3????21??23????21?????文章來源地址http://www.zghlxwxcb.cn/news/detail-406350.html
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