3.1 决定因素及其特性

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  决定因素及其特性

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  So far we have only learned about determinants when it came to discussing whether a matrix is invertible or not. Now, let's introduce the idea of a determinant more rigorously.
  ::到目前为止,我们只是在讨论矩阵是否不可置疑时才知道决定因素。 现在,让我们更严格地提出决定因素的概念。

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  The determinant of an  $\begin{align*}n\times n\end{align*}$   dimensional matrix of the form  $\begin{align*}A=[a_{ij}]\end{align*}$  is the summation of the form 
  ::窗体 A=[aij] 的 nxn 维矩阵的决定因素是窗体的总和

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  $\begin{align*}\det(A) = \sum_{j=1}^{n}(-1)^{j+1}a_{1j}\det(A_{1j})\end{align*}$
  ::det( A) =nj=1( 1) j+1a1jdet( A1j)

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  The terms of the form
  ::形式 形式 条件

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  $\begin{align*}C_{ij} = (-1)^{i+j}\det(A_{ij})\end{align*}$  are called cofactors
  ::Cij= (- 1) i+jdet (Aij) 被称为共构因素

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  so another form of the determinant can be written as
  ::因此,另一种形式的决定因素可以写成:

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  $\begin{align*}\det(A) = \sum_{j=1}^{n}a_{1j}C_{1j}\end{align*}$
  ::det( A) =nj=1a1jC1j

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  This is not just limited to cofactors of row 1, we can generalize the formal definition of the determinant to
  ::这不仅仅限于第1排的共同因素,我们可以将决定因素的正式定义概括到

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  $\begin{align*}\det(A)=\sum_{j=1}^{n}(-1)^{i+j}a_{ij}\det(A_{ij}) \\ \det(A)=\sum_{j=1}^{n}C_{ij}a_{ij}\end{align*}$
  ::det( A) =nj=1( 1) i+jaijdet( Aij) det( A) =nj=1Cijaij

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  Let's try an example of this:
  ::让我们试一试这个例子:

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  Calculate the determinant of
  ::计算

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  $\begin{align*}A = \begin{bmatrix}1&3&-4\\0&2&0\\-5&1&1\end{bmatrix}\end{align*}$
  ::A=[13-4020-511]

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  There are three ways to calculate this determinant so let's try each of them and prove we'll get the same result.
  ::有三种方法来计算这个决定因素 所以让我们试一下每一个 证明我们会得到同样的结果

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  $\begin{align*}\det(A) = \det\Bigg(\begin{bmatrix}1 & 3&-4\\0&2&0\\-5&1&1\end{bmatrix}\Bigg) \\ \det(A) = 1\det\Big(\begin{bmatrix}2&0\\1&1\end{bmatrix}\Big)-3\det\Big(\begin{bmatrix}0&0\\-5&1\end{bmatrix}\Big) + (-4)\det\Big(\begin{bmatrix}0&2\\-5&1\end{bmatrix}\Big) \\ \det(A) = 1(2) - 3(0) + (-4)(10) \\ \det(A) = 2-40 \\ \det(A) = -38\end{align*}$
  ::de(A)=det([13-4020-511])det(A)=1det([2011])-3det([00-51])+(-4)det([02-51])det(A)=1(2)-3(0)+(-4)(4)(10)det(A)=2-40det(A)______________________________________________

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  We can also solve this using the cofactors from the second row
  ::我们也可以用第二排的共生因素解决这个问题

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  $\begin{align*}\det(A) = \det\Bigg(\begin{bmatrix}1&3&-4\\0&2&0\\-5&1&1\end{bmatrix}\Bigg) \\ \det(A) = 0\det\bigg(\begin{bmatrix}3&-4\\1&1\end{bmatrix}\bigg) + 2\det\bigg(\begin{bmatrix}1&-4\\-5&1\end{bmatrix}\bigg) + 0\det\bigg(\begin{bmatrix}1&3\\-5&1\end{bmatrix}\bigg) \\ \det(A) = (0)(7)+2(-19)+0(16) \\ \det(A) = -38\end{align*}$
  ::de(A)=det([13-4020-511])det(A)=0det([3-411])+2det([1-4-51])+0det([13-51])det(A)=(0(7)+2(19)+0(16)det(A)____________________________________________________

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  So we get the same result and let's then finally do this for the third row
  ::所以结果是一样的,我们最后为第三排做这个

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  $\begin{align*}\det(A) = \begin{bmatrix}1&3&-4\\0&2&0\\-5&1&1\end{bmatrix} \\ \det(A) = -5\bigg(\begin{bmatrix}3&-4\\2&0\end{bmatrix}\bigg) -1\bigg(\begin{bmatrix}1&-4\\0&0\end{bmatrix}\bigg) -4\bigg(\begin{bmatrix}1&3\\0&2\end{bmatrix}\bigg) \\ \det(A) = -5(8) +1(0)+2 \\ \det(A) = -38\end{align*}$
  :A)=[13-4020-511]det(A)5([3-420])-1([1-400])-4([1302])det(A)5(8)+1(0)+2det(A)________________________________

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  So we see that this all works out. Now, what we can see is that the second row is easiest to do because it has the most 0's. So as a general rule of thumb, find which row is easiest to do cofactors with and then use that row.
  ::因此,我们看到这一切都可行。现在,我们可以看到第二行最容易做,因为它有最多0的缩略图。因此,作为一般的拇指规则,找到哪个行最容易做共犯,然后用那行。

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  Now, let's look at some other properties of the determinant.
  ::现在,让我们来看看其他的 决定因素的属性。

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  Let's look at triangular matrices to start off.
  ::让我们先看三角矩阵。

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  Let  $\begin{align*}A = \begin{bmatrix}2&0&0\\3&1&0\\1&5&-4\end{bmatrix}\end{align*}$
  ::让我们A=[20031015-4]

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  $\begin{align*}\det(A) = \det\Bigg(\begin{bmatrix}2 & 0&0\\3&1&0\\1&5&-4\end{bmatrix}\Bigg) \\ \det(A) = 2\det\bigg(\begin{bmatrix}1&0\\5&-4\end{bmatrix}\bigg) - 0\det\bigg(\begin{bmatrix}3&0\\1&-4\end{bmatrix}\bigg) + 0\det\bigg(\begin{bmatrix}3&1\\1&5\end{bmatrix}\bigg) \\ \det(A) = 2(-4) \\ \det(A) = -8 \end{align*}$  
  ::de(A) = det([20031015-44] det(A) = 2det([105-4]) - 0det([301-4]) + 0det([3115] det(A) = 2(-4) det(A) 8

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  We can actually see that the determinant of this lower triangular matrix is the product of the diagonals.
  ::实际上,我们可以看到,这种三角三角下方矩阵的决定因素是对角体的产物。

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  Now, let's prove this generally for any type of triangular matrix:
  ::现在,让我们为任何类型的三角矩阵 证明一下:

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  $\begin{align*}A = \begin{bmatrix}0&0&A_{13}\\0&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix} \\ \det(A) = 0\det\bigg(\begin{bmatrix}0&A_{22}\\A_{31}&A_{32}\end{bmatrix}\bigg) - 0\det\bigg(\begin{bmatrix}0&A_{23}\\A_{31}&A_{33}\end{bmatrix}\bigg) + A_{13}\det\bigg(\begin{bmatrix}0&A_{22}\\A_{31}&A_{32}\end{bmatrix}\bigg) \\ \det(A) = A_{13}\det\bigg(\begin{bmatrix}0&A_{22}\\A_{31}&A_{32}\end{bmatrix}\bigg) \\ \det(A) = A_{13}(0\cdot A_{32}-A_{22}A_{31}) \\ \det(A) = -A_{13}A_{22}A_{31}\end{align*}$  
  ::A=[00A130A22A23A23A31A32A33]det(A)=0det([0A22A31A32])-0det([0A23A31A33])+A13det([0A22A31A32]))det(A)=A13det([0A22A31A32])det(A)=A13(0QA32-A22A31)det(A) @A13A22A31

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  $\begin{align*}A = \begin{bmatrix}A_{11}&0&0\\A_{21}&A_{22}&0\\A_{31}&A_{32}&A_{33}\end{bmatrix} \\ \det(A) = \det\Bigg(\begin{bmatrix}A_{11}&0&0\\A_{21}&A_{22}&0\\A_{31}&A_{32}&A_{33}\end{bmatrix}\Bigg) \\ \det(A) = A_{11}\begin{bmatrix}A_{22}&0\\A_{32}&A_{33}\end{bmatrix} + 0\cdots + 0\cdots \\ \det(A) = A_{11}(A_{22}A_{33} - 0) \\ \det(A) = A_{11}A_{22}A_{33}\end{align*}$  
  ::A=[A1100A21A220A220A31A32A33]det(A)=det([A1100A21A220A220A31A32A33])det(A)=A11[A220A32A33]+00det(A)=A11(A22A33-0)det(A)=A11A22A33

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  So for a right upper triangular matrix and a left upper triangular matrix we have that the determinants are both the product and the negative product of the determinant. 
  ::因此,右上三角矩阵和左上三角矩阵的决定因素既是决定因素的产物,也是其负面产物。

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  Next, we follow with a similar proof for a lower right and lower left triangular matrix to get 
  ::接下来,我们用类似的证据 来证明右下右下左下三角矩阵

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  $\begin{align*}A=\begin{bmatrix}A_{11} & A_{12}&A_{13}\\0&A_{22}&A_{23}\\0&0&A_{33}\end{bmatrix} \\ \det(A) = A_{11}\det\bigg(\begin{bmatrix}A_{22}&A_{23}\\0&A_{33}\end{bmatrix}\bigg) - A_{12}\det\bigg(\begin{bmatrix}0&A_{23}\\0&A_{33}\end{bmatrix}\bigg) + A_{13}\det\bigg(\begin{bmatrix}0&A_{22}\\0&0\end{bmatrix}\bigg) \\ \det(A) = A_{11}(A_{22}A_{33}-0\cdot A_{23}) - A_{12}(0A_{33}-0A_{23}) + A_{13}(0-A_{22}0) \\ \det(A) = A_{11}A_{22}A_{33}\end{align*}$
  ::A=[A11A12A12A130A22A2300A33]det(A)=A11det[[A22A230A33])-A12det([0A230A33])+A13det([0A2200])det(A)=A11(A22A3-3-001A23)-A12(0A33-0A223)+A13(0-A220)dt(A)=A11A22A33)

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  Now we'l go on to the next triangular matrix
  ::现在我们继续下一个三角矩阵

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  $\begin{align*}A = \begin{bmatrix}A_{11} & A_{12}&A_{13}\\A_{21}&A_{22}&0\\A_{31}&0&0\end{bmatrix} \\ \det(A) = \det\Bigg(\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&0\\A_{31}&0&0\end{bmatrix}\Bigg) \\ \det(A) = A_{11}\det\bigg(\begin{bmatrix}A_{22}&0\\0&0\end{bmatrix}\bigg) - A_{12}\det\bigg(\begin{bmatrix}A_{21}&0\\A_{31}&0\end{bmatrix}\bigg) + A_{13}\det\bigg(\begin{bmatrix}A_{21}&A_{22}\\A_{31}&0\end{bmatrix}\bigg) \\ \det(A) = 0 - 0 + A_{13}(0-A_{22}A_{31}) \\ \det(A) = -A_{13}A_{22}A_{31} \end{align*}$
  ::A=[A11A12A1313A21A220A3100]det(A)=det([A11A12A13A13A21A21A220A3100])det(A)=A11det[[A2000])-A12det([A210A310]))+A13det([[A21A22A310]))det(A)=0-0+A13(0-A22A31)det(A)_A13A22A31

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  Now, that's look at one more property of determinants, multiplicative-ness and inverses.
  ::现在,这是再看一个决定因素的属性, 多重性和反向性。

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  First, let's look at determinants of elementary matrices and determinants of powers and inverses of matrices.
  ::首先,让我们看看基本矩阵的决定因素, 以及权力的决定因素和矩阵的反面。

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  The determinant of an elementary matrix times another matrix A is the product of the determinants, so if $\begin{align*}E\end{align*}$  is an elementary matrix and
  ::基本矩阵的决定因素乘以另一个矩阵A是决定因素的产物,因此如果E是基本矩阵的话,

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  $\begin{align*}A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix} \\ \det(EA) = \det(E)\det(A) \end{align*}$
  ::A=[A11A12A12A21A22]det(EA)=det(E)det(A)

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  Now we can gather that
  ::现在我们可以收集到

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  $\begin{align*}\det(AB) = \det(A)\det(B) \\ \text{Because... } \\\ A = E_{1}E_{2}\cdots E_{n} \\ \det(E_{1}E_{2}\cdots E_{n}B) \\ = \det(E_{1})\det(E_{2})\cdots\det(E_{n})\det(B) \\ = \det(E_{1}E_{2}\cdots E_{n})\det(B) \\ = \det(A)\det(B) \end{align*}$
  ::de( AB) = det( A) ded( B) 因为... A= E1E2 _Endet( E1E2 @EnB) = det( E1E2) dedt( E2) {det( E2) adt( B) = det( E1E2E) det( B) adt( B) ditt( B) = det( A) dedt( B)

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  Similarly, for inverses
  ::同样,对于逆数

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  $\begin{align*}AA^{-1} = I \\ \det(AA^{-1}) = \det(A)\det(A^{-1}) \\ \det(A)\det(A^{-1}) = \det(I) \\ \det(A)\det(A^{-1}) = 1 \\ \det(A) = \frac{1}{\det(A^{-1})}\end{align*}$
  ::A- 1 = Idet (A- 1) = det (A-1) det (A-1) det (A-1) det (A-1) det (A-1) det (A-1) det (A-1) det (A-1) idet (A-1) = 1det (A-1) = 1det (A-1)

   

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