节: 矩阵反转和共构件扩展 | 3.3 矩阵反转和共构体扩展

章节大纲

We know how to take the inverse of a 2x2 matrix, but let's move on and see how cofactors and determinants play in taking inverses of higher dimensional matrices.
::我们知道如何逆向 2x2 矩阵, 但让我们继续看共生因素和决定因素如何在反向高维矩阵中起作用。

We know what the cofactors are from each of our last few lessons. So, let's define what the cofactor matrix is.
::我们知道我们最后几堂课的共生因素是什么。让我们来定义一下共生因素矩阵是什么。

Given an $\begin{aligned} n \times n \end{aligned}$ matrix A, the cofactor matrix of A, denoted by C, can be written as the matrix where each entry is equivalent to the cofactor in matrix A.
::鉴于 nxn 矩阵A,C 表示的A的共构物矩阵可以写成矩阵,其中每个条目相当于矩阵A中的共构物。

Let's try and visualize this with an example.
::让我们用一个例子来想象一下。

$\begin{aligned} A = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 4 & - 2 \\ 0 & 5 & - 3 \end{matrix}] \\ c_{11} = det ([\begin{matrix} 4 & - 2 \\ 5 & - 3 \end{matrix}]), c_{12} = - det ([\begin{matrix} 3 & - 2 \\ 0 & - 3 \end{matrix}]), c_{13} = det ([\begin{matrix} 3 & 4 \\ 0 & 5 \end{matrix}]) \\ c_{21} = - det ([\begin{matrix} - 1 & 2 \\ 5 & - 3 \end{matrix}]), c_{22} = det ([\begin{matrix} 1 & 2 \\ 0 & - 3 \end{matrix}]), c_{23} = - det ([\begin{matrix} 1 & - 1 \\ 0 & 5 \end{matrix}]) \\ c_{31} = det ([\begin{matrix} - 1 & 2 \\ 4 & - 2 \end{matrix}]), c_{32} = - det ([\begin{matrix} 1 & 2 \\ 3 & - 2 \end{matrix}]), c_{33} = det ([\begin{matrix} 1 & - 1 \\ 3 & 4 \end{matrix}]) \\ C = [\begin{matrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{matrix}] \\ C = [\begin{matrix} | \begin{matrix} 4 & - 2 \\ 5 & - 3 \end{matrix} | & - | \begin{matrix} 3 & - 2 \\ 0 & - 3 \end{matrix} | & | \begin{matrix} 3 & 4 \\ 0 & 5 \end{matrix} | \\ - | \begin{matrix} - 1 & 2 \\ 5 & - 3 \end{matrix} | & | \begin{matrix} 1 & 2 \\ 0 & - 3 \end{matrix} | & - | \begin{matrix} 1 & - 1 \\ 0 & 5 \end{matrix} | \\ | \begin{matrix} - 1 & 2 \\ 4 & - 2 \end{matrix} | & - | \begin{matrix} 1 & 2 \\ 3 & - 2 \end{matrix} | & | \begin{matrix} 1 & - 1 \\ 3 & 4 \end{matrix} | \end{matrix}] \\ C = [\begin{matrix} - 2 & 9 & 15 \\ 7 & - 3 & - 5 \\ - 6 & 8 & 7 \end{matrix}] \end{aligned}$
::A=[1-1234-205-33],c11=det([4-25-3],c12det([3-20-3]),c13=det([3405])c21det([125-3]),c22=det([120-3]),c23det([1-105])c31=det([-124-2],c32det([123-2]),c33=det([1-134])C=[c11c12c13c21c22c-23c31c3233] C=[425-32______________________________________________________________________________________________(120_____)c23_______________________________________________________________________________________([291-1573-3_________687]

And using this matrix, if we take it's transpose and divide by the determinant of A, we get the inverse of A. So let's check that
::使用这个矩阵,如果我们把它转换和除以A的决定因素, 我们得到A的反反。让我们检查一下

$\begin{aligned} C = [\begin{matrix} - 2 & 9 & 15 \\ 7 & - 3 & - 5 \\ - 6 & 8 & 7 \end{matrix}] \\ C^{T} = [\begin{matrix} - 2 & 7 & - 6 \\ 9 & - 3 & 8 \\ 15 & - 5 & 7 \end{matrix}] \\ det (A) = - 2 (1) - 1 (9) + 2 (15) \\ det (A) = 19 \\ A^{- 1} = \frac{1}{19} [\begin{matrix} - 2 & 7 & - 6 \\ 9 & - 3 & 8 \\ 15 & - 5 & 7 \end{matrix}] \\ A^{- 1} = [\begin{matrix} - \frac{2}{19} & \frac{7}{19} & - \frac{6}{19} \\ \frac{9}{19} & - \frac{3}{19} & \frac{8}{19} \\ \frac{15}{19} & - \frac{5}{19} & \frac{7}{19} \end{matrix}] \end{aligned}$
::C=[29157-3-3-3-5-5-687]C=[-27-69-3815-57]det(A) 2(1)-1(9)+2(15)=19A-1=119[-27-38-15-57]A-1=219719-619919-31981919-19-19-519719]

Another name for the cofactor matrix transposed is the adjugate or adjunct matrix.
::移植的共构矩阵的另一个名称是辅助矩阵或辅助矩阵。

So, $\begin{aligned} A^{- 1} = \frac{1}{det (A)} adj (A) \end{aligned}$

::所以,A-1=1det(A)adj(A)

Similarly, from this we can also gather the property that $\begin{aligned} (adj (A)) \cdot A = det (A) \cdot I \end{aligned}$ where again, $\begin{aligned} I \end{aligned}$ is the identity matrix.
::同样,我们也可以从中收集(adj(A))_A=det(A)_I的财产,而I是身份矩阵。

So we can prove this by seeing that
::因此,我们可以通过看到这一点来证明这一点:

$\begin{aligned} A^{- 1} = \frac{1}{det (A)} adj (A) \\ A \cdot A^{- 1} = A \cdot \frac{1}{det (A)} adj (A) \\ I = A \cdot \frac{1}{det (A)} adj (A) \\ det (A) = A adj (A) \\ A (adj (A)) = det (A) \cdot I \end{aligned}$
::A-1=1det(A)adj(A)AQA-1=AQ1det(A)AA(A)I=AQIdet(A)Aadj(A)det(A)A(A)adj(A)A(A)A(A)=det(A)I

Now, let's verify this with an example
::现在,我们举个例子来验证一下

$\begin{aligned} A = [\begin{matrix} 1 & 2 & - 1 \\ 0 & 4 & 0 \\ 3 & - 2 & 1 \end{matrix}] \\ det (A) = 16 \\ C = [\begin{matrix} 4 & 0 & - 12 \\ 0 & 4 & 0 \\ 4 & 0 & 4 \end{matrix}] \\ C^{T} = [\begin{matrix} 4 & 0 & 4 \\ 0 & 4 & 0 \\ - 12 & 0 & 4 \end{matrix}] \\ adj (A) = [\begin{matrix} 4 & 0 & 4 \\ 0 & 4 & 0 \\ - 12 & 0 & 4 \end{matrix}] \\ A (adj (A)) = [\begin{matrix} 1 & 2 & - 1 \\ 0 & 4 & 0 \\ 3 & - 2 & 1 \end{matrix}] \cdot [\begin{matrix} 4 & 0 & 4 \\ 0 & 4 & 0 \\ - 12 & 0 & 4 \end{matrix}] \\ [\begin{matrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{matrix}] \\ = 16 I \end{aligned}$
::A=[12-104003-21]det(A)=16C=[40-12040404]CT=[404040-12004]adj(A)=[404040-12004]A(adj(A))=[12-104003-21]__[404040-1204][16006006]=16I

Now, let's look at products of inverses
::现在,让我们看看逆向的产品

Let $\begin{aligned} C = A B \end{aligned}$ where A and B are square matrices both with dimension $\begin{aligned} n \times n \end{aligned}$ .
::在 A 和 B 是方格的 C= AB 中, 以维度 nxn 表示。

Hence
::因此

$\begin{aligned} C = A B \\ A^{- 1} C = A^{- 1} A B \\ A^{- 1} C = B \\ B = A^{- 1} C \\ C B^{- 1} = A B B^{- 1} \\ A = C B^{- 1} \\ C = A B \\ C = (C B^{- 1}) (A^{- 1} C) \\ C = C B^{- 1} A^{- 1} C \\ C C^{- 1} = C B^{- 1} A^{- 1} C C^{- 1} \\ I = C B^{- 1} A^{- 1} \\ C^{- 1} = C^{- 1} C B^{- 1} A^{- 1} \\ C^{- 1} = I B^{- 1} A^{- 1} \\ C^{- 1} = B^{- 1} A^{- 1} \end{aligned}$
::C=ABA-1C=A-1A-1ABA-1C=A-1ABB=A-1CB-1C=A-1CB-1A=ABB-1C-1A=ABB-1A=ABB-1C=CB-1C=(CB-1)(A-1C)C=CB-1A-1A-1CCC-1C-1C=CB-1A-1A-1C-1C-1C-1C=CB-1A-1C-1C-1C-1C-1C-1C-1C-1C-1C-1=CB-1A-1A-1A-1A-1C-1B-1B-1C-1=B-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1=CB-1A-1A-1A-1A-1C-1A-1A-1C-1C-1C-1C-1C-1C-1C-1=B-1B-1C-1B-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1=C-1C-1C-1C-1C-1C-1C-1=C-1=C-1=C-1=C-1C-1C-1C-1C-1C-1C-1=C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1=C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1=C-1C-1C-1=C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1(C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C-1C

We'll go into some matrix inversion techniques later in this book, but we can introduce a few other matrix inversion techniques now.
::我们可以在这本书的后期采用一些矩阵反向技术但我们现在可以引进其他一些矩阵反向技术

Another way to find the inverse of a matrix is to use Gauss Jordan elimination
::另一种发现矩阵反反比的方法是使用约旦消除高斯的方法。

In order to solve this, we write
::为了解决这个问题,我们写

$\begin{aligned} [A | I] \end{aligned}$ and then row reduce to make it into the form $\begin{aligned} [I | A^{- 1}] \end{aligned}$
::[AI],然后排行缩小,以形成表格[IA-1]

and then this is true, because if you look at how we solve the system $\begin{aligned} A \vec{x} = \vec{b} \end{aligned}$ and write the system $\begin{aligned} [A | \vec{b}] \end{aligned}$ and then get $\begin{aligned} [I | A^{- 1} \vec{b}] \end{aligned}$
::然后,这是真的, 因为如果你看看我们如何解决系统 Axb 和写系统[Ab] 然后得到[IA-1b]

So we can see that solving for A inverse is solving the equation $\begin{aligned} A B = I \end{aligned}$ and trying to find the matrix $\begin{aligned} B \end{aligned}$ .
::因此,我们可以看到,解决反向的方法就是解决AB=I的方程式, 并试图找到矩阵B。

Let's look at an example of this:
::举个例子:

$\begin{aligned} A = [\begin{matrix} 1 & - 2 \\ - 3 & 5 \end{matrix}] \\ [A | I] = [\begin{matrix} 1 & - 2 & | & 1 & 0 \\ - 3 & 5 & | & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 1 & - 2 & | & 1 & 0 \\ 0 & - 1 & | & 3 & 1 \end{matrix}] \\ = [\begin{matrix} 1 & 0 & | & - 5 & - 2 \\ 0 & - 1 & | & 3 & 1 \end{matrix}] \\ = [\begin{matrix} 1 & 0 & | & - 5 & - 2 \\ 0 & 1 & | & - 3 & - 1 \end{matrix}] \end{aligned}$
::A=[1-2-2-35][AI]=[1-2][10-3501]=[1-2____________________________________________________________________________[10_____________________________________________________________________________________________[10________________________________________________[1________________________________________________________[10____________________________________[10____________________________________[10________________________________________________________________________________________________________________[[[[[10______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[[[[[[[[[[[[[[[[________________________________________________________________________________________________

And $\begin{aligned} A^{- 1} = [\begin{matrix} - 5 & - 2 \\ - 3 & - 1 \end{matrix}] \end{aligned}$
::A-1=[-5-2-3-1]

To verify that this is true we see that
::为了核实事实真相,我们看到:

$\begin{aligned} [\begin{matrix} 1 & - 2 \\ - 3 & 5 \end{matrix}] \cdot [\begin{matrix} - 5 & - 2 \\ - 3 & - 1 \end{matrix}] \\ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I \end{aligned}$
::[1-2-35]-[-5-2-3-3-1]=[1001]=一