得鹿梦鱼 得鹿梦鱼

逆矩阵

设给定一个线性变换

{y1=a11x1+a12x2+...+a1nxny2=a21x1+a22x2+...+a2nxn...yn=an1x1+an2x2+...+annxn\begin{cases}y_1 = a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n \\y_2 = a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n \\... \\y_n = a_{n1}x_1 + a_{n2}x_2 + ... + a_{nn}x_n \\\end{cases}

它的系数矩阵式一个n阶矩阵A\bold A
X=[x1x2...xn]y=[y1y2...yn]\bold X =\left [\begin {matrix}x_1 \\x_2 \\... \\x_n\end {matrix}\right ]\text{, }\bold y =\left [\begin {matrix}y_1 \\y_2 \\... \\y_n\end {matrix}\right ]
则上述变换可以记作
AX=Y1\bold A \bold X = \bold Y \tag{1}
A\bold A的伴随矩阵A\bold A^*左乘上式两端
AAX=AY\bold A^* \bold A \bold X = \bold A^* \bold Y
AEX=AY\bold A\bold E \bold X = \bold A^* \bold Y
AX=AY\bold A\bold X = \bold A^* \bold Y
A0\bold A \neq 0,可解出
X=1AAY\bold X =\frac{1}{\bold A}\bold A^* \bold Y

B=1AA\bold B =\frac{1}{\bold A}\bold A^*
X=BY2\bold X = \bold B \bold Y \tag{2}
2\text{2}式表示一个从Y到X的线性变换,称为1\text{1}式的逆变换

对于一个n阶矩阵A,如果有一个n阶矩阵B,使得
AB=BA=E\bold A \bold B = \bold B \bold A = \bold E
则说矩阵A是可逆的,并把B称为A的逆矩阵,记作A1\bold A^{-1}逆矩阵是唯一的

定理1: 若矩阵A可逆,则A0\bold A \neq 0
定理2: 若A0\bold A \neq 0,则矩阵A可逆,且A1=1AA\bold A^{-1} = \frac{1}{\bold A} \bold A^*

A0\bold A \neq 0时,A\bold A称为奇异矩阵,否则被成为非奇异矩阵

  1. 若矩阵可逆,A11=A\bold A^{-1}^{-1} = \bold A
  2. 若矩阵可逆且λ0,λA\lambda \neq 0, \lambda \bold A可逆,λA1=1λA1\lambda \bold A^{-1}= \frac{1}{\lambda}\bold A^{-1}
  3. AB为同阶矩阵且均可逆AB1=B1A1\bold A \bold B^{-1} = \bold B^{-1} \bold A^{-1}
  4. A可逆,且AT\bold A^T也可逆,则AT1=A1TA^T^{-1} = \bold A^{-1}^T
  5. A可逆, a,b为正数使,AaAb=Aa+b,Aab=Aab\bold A^a\bold A^b = \bold A^{a + b}, \bold A^a^b = \bold A^{ab}

速算二阶矩阵的逆矩阵

[abcd]\left [\begin{matrix}a & b \\c & d\end{matrix}\right]

则逆矩阵为
1adbc[dbca]\frac{1}{ad -bc}\left [\begin{matrix}d & -b \\-c & a\end{matrix}\right]