Course: 3.2 决定因素的更多属性

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更多的决定因素特性
In this lesson we will learn about more relationships with matrices and their determinants.
::在这一教训中,我们将了解与矩阵及其决定因素的更多关系。

If the matrix $\begin{aligned} B \end{aligned}$ is constructed by adding a scalar multiple of one row of $\begin{aligned} A \end{aligned}$ to another row of $\begin{aligned} A \end{aligned}$ we get that $\begin{aligned} det (A) = det (B) \end{aligned}$

As an example of this we see that
::作为这方面的一个例子,我们看到:

$\begin{aligned} A = [\begin{matrix} 1 & 2 \\ - 3 & 1 \end{matrix}] \\ B = [\begin{matrix} 1 & 2 \\ - 1 & 5 \end{matrix}] \end{aligned}$
::A=[12-31]B=[12-15]

This is constructed by $\begin{aligned} 2 R_{1} + R_{2} \to R_{2} \end{aligned}$ to get the second matrix
::这是由 2R1+R2QR2 构造的, 以获得第二个矩阵

$\begin{aligned} det (A) = det ([\begin{matrix} 1 & 2 \\ - 3 & 1 \end{matrix}]) \\ det (A) = 1 (1) - 2 (- 3) = 1 - (- 6) = 7 \\ det (B) = det ([\begin{matrix} 1 & 2 \\ - 1 & 5 \end{matrix}]) \\ det (B) = 5 (1) - 2 (- 1) = 5 + 2 = 7 \\ det (A) = det (B) \end{aligned}$
::de( A) =det( [12- 31] =det( A) =1(1)- (2- 3) =1- (-6) = 7det( B) =det( [12- 15) det( B) =5(1)- (-2) - 1) =5+2=7det( A) =det( B)

::如果矩阵B的构造方法是在另一行A中添加一行AA的斜体倍数,我们就会得到该 det(A)=det(B) 作为这方面的一个例子,我们看到A=[12-31]B=[12-15]B=[12-31]B=[12-1+R2-15],这是用 2R1+R2=(A)=det([12-31])det(A)=1-(1)-3)=1-(-6)=7det(B)=det([12-15])det(B)=5(1)-(2(-2)(-1)=5+2=7det(A)=det(B)

If the matrix $\begin{aligned} B \end{aligned}$ is constructed by swapping two rows of $\begin{aligned} A \end{aligned}$ then we get that $\begin{aligned} det (A) = - det (B) \end{aligned}$

As an example let's show that
::举例来说,让我们来证明

$\begin{aligned} A = [\begin{matrix} 3 & - 1 \\ - 2 & 5 \end{matrix}] \\ B = [\begin{matrix} - 2 & 5 \\ 3 & - 1 \end{matrix}] \\ det (A) = det ([\begin{matrix} 3 & - 1 \\ - 2 & 5 \end{matrix}]) \\ det (A) = 3 (5) - (- 1) (- 2) \\ det (A) = 15 - 2 \\ det (A) = 13 \\ det (B) = det ([\begin{matrix} - 2 & 5 \\ 3 & - 1 \end{matrix}]) \\ det (B) = - 2 (- 1) - 5 (3) \\ det (B) = 2 - 15 \\ det (B) = - 13 \\ det (A) = - det (B) \end{aligned}$
::A=[3-1--25]B=[-253-1-1-1]det(A)=det([3-1-25])det([3-1-25])det(A)=-3(5)-(-1)(-2)(-2)det(A)=15-2det(A)=13det(B)=det([-253-1)det(B)(2-1)-5(3)det(B)=2-15det(B) 13det(A)\det(B)

::如果矩阵B是通过对 A 的两行进行交换而构造的,那么我们就可以获得 det( A) {%det( B) 作为例子,让我们以A=[ 3- 1- 25] B=[-253-1-1-1)det( A) =det ([ 3-1- 25]) det( A) =3(5)- (- 1) (A) = 15-2det( A) =13det( B) =det( [-253-1) det( B) =2- 15det( B) =2- 15det( B) {13det( A) =13-2det( B)

Lastly, if $\begin{aligned} B \end{aligned}$ is constructed by multiplying one row of $\begin{aligned} A \end{aligned}$ by a scalar $\begin{aligned} c \end{aligned}$ we'll end up getting that $\begin{aligned} det (A) = \frac{1}{c} det (B) or det (B) = c det (A) \end{aligned}$

As an example,
::例如,

$\begin{aligned} A = [\begin{matrix} 3 & - 4 \\ 2 & 5 \end{matrix}] \\ B = [\begin{matrix} 9 & - 12 \\ 2 & 5 \end{matrix}] \\ det (A) = 3 (5) - 2 (- 4) \\ det (A) = 15 + 8 \\ det (A) = 23 \\ det (B) = 9 (5) - 2 (- 12) \\ det (B) = 45 + 24 \\ det (B) = 69 \\ det (B) = 3 det (A) \\ det (A) = \frac{1}{3} det (B) \end{aligned}$
::A=[3-425]B=[9-1225]B=[9-1225]det(A)=[9-1225)det(A)=[3-1225)det(A)=[3-1225)det(A)=[15+8det(A)=(B)=23det(B)=(9)(5)-2(-2)-(-12)det(B)=45+24det(B)=69dt(B)=(A)det(A)det(A)=13det(B)

To get $\begin{aligned} B \end{aligned}$ multiply row 1 of $\begin{aligned} A \end{aligned}$ by 3
::获得 B 乘以 A 1 乘以 A 乘以 3

::最后,如果B的构造是乘以一行A乘以一行A乘以一个scalar c,我们最终将获得det(A)=1cdet(B)或det(B)=cdet(A)作为例子,A=[3-425,B=[9-1225,B=[9-1225,B=[A)=3(5)(A)=3(5)-2-4(2-4)det(A)=15+8det(A)=23dt(B)=9(5)-(2-(B)=45+24det(B)=69det(B)=3det(A)det(A)=13det(B) 乘A第一行x3

Let's look at another matrix property for determinants

Given a matrix $\begin{aligned} A \end{aligned}$ the determinant of the transpose of it is equal to its determinant, in other words, $\begin{aligned} det (A) = det (A^{T}) \end{aligned}$

::相对于其决定因素,即det(A)=det(AT),即det(A)=det(AT)

Let's try and verify this for ourselves

$\begin{aligned} A = [\begin{matrix} - 4 & 1 \\ 3 & 2 \end{matrix}] \\ A^{T} = [\begin{matrix} - 4 & 3 \\ 1 & 2 \end{matrix}] \\ det (A) = (- 4) (2) - 3 (1) \\ det (A) = - 8 - 3 \\ det (A) = - 11 \\ det (A^{T}) = - 4 (2) - 3 \\ det (A^{T}) = - 11 \\ det (A) = det (A^{T}) \end{aligned}$
::A=[-4132]AT=[-4312]det(A)=(-4)(2)-3(1)det(A) 8-3det(A) 11det(AT) 4(2)-3det(AT) 11det(A) =det(AT)

::让我们尝试为我们自己验证A=[-4132]AT=[-4312]det(A)=(-4)(2)-3(1)det(A)8-3det(A)11det(AT)4(2)-3det(AT)11det(A)=(A)det(AT)

Now, let's try to prove it

To prove this let's look at the principal of mathematical induction. The principal of mathematical induction states that if the statement "> $\begin{aligned} P (n) \end{aligned}$ is true at the base case of $\begin{aligned} P (1) \end{aligned}$ and then assuming $\begin{aligned} P (k) \end{aligned}$ is true then if $\begin{aligned} P (k + 1) \end{aligned}$ is true then "> $\begin{aligned} P (n) \end{aligned}$ is true for every integer.
::为了证明这一点,让我们看看数学感应的原理。数学感应的原理指出,如果语句P在P(1)的基本情况中是真实的,然后假设P(k)是真实的,那么,如果P(k+1)是真实的,那么P在每个整数中都是真实的。

So, we know that this statement is true for a 1x1 matrix, so let's assume that this statement is true for a k x k matrix
::因此,我们知道,对于1x1矩阵来说,这一声明是真实的,因此,让我们假设,对于 k x k 矩阵来说,这一声明是真实的。

In $\begin{aligned} A \end{aligned}$ , the cofactor of $\begin{aligned} a_{1 j} \end{aligned}$ is the cofactor $\begin{aligned} a_{j 1} \end{aligned}$ in $\begin{aligned} A^{T} \end{aligned}$ so they have cofactor expansions which are equal because the cofactor expansion down rows of A is the cofactor expansion down the columns of the transpose of A.
::在A中,a1j的同源物是AT的同源物aj1,因此它们具有相等的同源物膨胀,因为A下行的同源物膨胀是A转基因柱下的同源物膨胀。

That, in actuality means that A and its transpose are equal so by induction we are done.
::这实际上意味着,通过上岗培训,A及其转换是平等的。

::现在,让我们来证明这一点。让我们来证明一下数学感应的原理。数学感应的原理指出, 如果语句 P 在P(1) 的基数中是真实的, 然后假设 P(k) 是真实的, 那么如果P( k+1) 是真实的, 那么P 在每个整数中都是真实的。因此, 我们知道这个语句对 1x1 矩阵来说是真实的, 所以让我们假设这个语句对 k x k 矩阵在 A 中是真实的, a1j 的共构体是AT 的共构体 aj1, 所以它们具有等同的共构体扩展, 因为 A 行下方的共构体扩展是 A 转换的列下方的共构体扩展。实际上, A 和它的转体是等同的, 我们完成的感应。

::让我们查看其它决定因素的矩阵属性。矩阵 A 转换的决定因素等于其决定因素。换句话说, 换句话说, 校对(AT) 让我们尝试为我们自己验证 A=[- 4132]AT=[- 4312] 代(A) = (4) (2) - 3 (A) (2) - 3 (det) (2) - 3det(A) = 3det (AT) 4\ 4(2) - 3det (AT) 4\ 4 (2) - 3det (AT) 。因此, 让我们试着证明这个声明对 k x\\\\\\ 11det (A) = adt(AT) 等于其决定因素。数学感官( AT) 让我们来证明这一点。数学感官( AT) 让我们来证明这一点。数学感官让我们来证明这个声明是真实的。数学感官 AT) AT atembortiquestrain is the real a developtional a cofacal a listrations.

Let's try out some new practice problems to see if we can apply these properties:

Prove that if $\begin{aligned} A \end{aligned}$ is an $\begin{aligned} n \times n \end{aligned}$ matrix and it has the property that $\begin{aligned} A^{T} A = I \end{aligned}$ , where $\begin{aligned} I \end{aligned}$ equals the identity matrix then $\begin{aligned} det (A) = \pm 1 \end{aligned}$

To prove this we get that
::为了证明这一点,我们得到

$\begin{aligned} det (A^{T} A) = det (I) \\ det (A^{T} A) = 1 \\ det (A^{T}) det (A) = 1 \\ det (A) = det (A^{T}) \\ det (A)^{2} = 1 \\ det (A) = \pm 1 \end{aligned}$

hence $\begin{aligned} det (A) = \pm 1 \end{aligned}$
::因此(A) @%1

::de(ATA) = det(I) det(AT) = 1det(AT) det(A) = 1det(A) = det(AT) det(A) 2 = 1det(A) = 1det(A) = 1 det(A) = 1 因而 det(A) = 1

::证明如果 A 是 nxn 矩阵, 并且它具有 ATA=I 的属性, 也就是我等同于身份矩阵的属性, 那么 det( A)\\\\ 1 来证明这一点, 我们得到的是 det( ATA) = det( I) dedet( AT) adet( A) = 1det( A) = 1det( A) = det( AT) ded( A) 2= 1det( A) = 1 dedet( A) 1 因而是 det( A)\\ 1

::让我们尝试一些新的实践问题, 看看我们是否可以应用这些属性 : 证明如果 A 是 nxn 矩阵, 并且它具有 ATA=I 的属性, 则我等于身份矩阵, 然后 det( A)\\\\\ 1 来证明我们得到了 det( ATA) =1det( AT) dedet( A) =1det( A) =1det( A) =det( AT) dedet( A) 2=1det( A) =1 dedet( A) 。

If
::如果(如果)

$\begin{aligned} A = [\begin{matrix} 1 & - 2 & 3 \\ - 1 & 5 & 7 \\ - 2 & 3 & 1 \end{matrix}] \\ det (A) = 31 \\ B = [\begin{matrix} - 1 & 8 & 3 \\ - 1 & 5 & 17 \\ - 2 & 3 & 1 \end{matrix}] \end{aligned}$
::A=[1-23-157-231]det(A)=31B=[-183-1517-231]

Calculate $\begin{aligned} det (B) \end{aligned}$

Look at this really closely, one can see that 2 times row two of matrix A added to row 1 and sent to row one creates matrix B. Hence, from our earlier rules we can gather that
::仔细看看这个,人们可以看到, 矩阵A的第2行2倍加在第1行, 并发送到第1行第1行, 创建了矩阵B。因此,根据我们早期的规则,我们可以收集到

$\begin{aligned} det (B) = det (A) \\ det (A) = 31 \\ \to det (B) = 31 \end{aligned}$
::de(B)=det(A)det(A)=31&det(B)=31

::计算 det( B) 仔细查看这个非常接近的值, 人们可以看到, 矩阵A 2 倍于 2 行 2 的矩阵A 添加到第 1 行, 并发送到第 1 行 1 创建矩阵 B 。因此, 根据我们先前的规则, 我们可以看到 dt( B) = det( A) dedet( A) = 31&det( B) =31

::如果A=[1-23-157-157-231]det(A)=31B=[-183-1517-2311] 计算 det(B) , 计算 det(B) 仔细看, 人们可以看到, 矩阵A 2 行的2倍 2 的矩阵A 添加到行1 , 并发送到行一创建矩阵 B。因此, 根据我们先前的规则, 我们可以看到 det( B) =det( A) adet( A) = 31det( B) =31

Calculate
::计算计算

$\begin{aligned} det (A B) \\ A = [\begin{matrix} 0 & 5 & 1 \\ - 3 & 5 & 2 \\ 1 & 2 & - 7 \end{matrix}] \\ B = [\begin{matrix} - 9 & 15 & 6 \\ 0 & 5 & 1 \\ 1 & 2 & - 7 \end{matrix}] \end{aligned}$

$\begin{aligned} A = [\begin{matrix} 0 & 5 & 1 \\ - 3 & 5 & 2 \\ 1 & 2 & - 7 \end{matrix}] \\ B = [\begin{matrix} - 9 & 15 & 6 \\ 0 & 5 & 1 \\ 1 & 2 & - 7 \end{matrix}] \\ det (A) = 0 det ([\begin{matrix} 5 & 2 \\ 2 & - 7 \end{matrix}]) - 5 det ([\begin{matrix} - 3 & 2 \\ 1 & - 7 \end{matrix}]) + 1 det ([\begin{matrix} - 3 & 5 \\ 1 & 2 \end{matrix}]) \\ det (A) = 0 (5 (- 7) - 2 (2)) - 5 (- 3 (- 7) - 2 (1)) + 1 (- 3 (2) - 5 (1)) \\ det (A) = 0 (- 35 - 4) - 5 (21 - 2) + 1 (- 6 - 5) \\ det (A) = 0 (- 39) - 5 (19) + 1 (- 11) \\ det (A) = 0 - 95 - 11 \\ det (A) = - 106 \end{aligned}$
::A=[051-352-212-7]B=[-9156-112-7]det(A)=0det([522-7])-5det([-321-1-7]))+1det([-321-7])+1det([[-3512])det([5-(7)-2(2))-(5)-(3-7)-2(1))+1(3(2)-5(1))-1(3(2)-5(1))det(A)=0(-35-4)-4)-5(21-2)+1(-6-5)det(A)=0(-39)-5(19)+1(-(11)det(A)=0-95-(A)_106

We can also see that from matrix B we have that rows one and two are swapped and then the new row one is multiplied by 3. Hence, we can gather that the determinant of B can be written as the product of the determinant of those elementary matrices times the determinant of matrix A
::我们还可以看到,从表B中我们可以看到,一行和二行被互换,然后新的一行乘以3。因此,我们可以发现,B的决定因素可以写成是这些基本矩阵的决定因素乘以矩阵A的决定因素的产物。

$\begin{aligned} det (B) = - 3 det (A) \\ det (A B) = det (A) det (B) \\ det (A) (- 3 det (A)) = - 3 det (A)^{2} \\ det (A) = - 106 \\ det (A B) = - 3 (- 106)^{2} \\ det (A B) = - 33708 \end{aligned}$
::d(B) 3det(A) adt(AB) = det(A) det(B) det(A)(A) 3det(A) 3det(A) 3det(A) 2det(A) 106det(AB) 3(106) 2det(AB) 3*33708

:A) = 0det([522-7])+1det([-351-1-7])+1det([-3512])+5(A)=[051-3(7)-2(2))-5(3-3(3-7)-2(1))+(3)-(2)-(2)-(3)-(5)-(5)-(A)=[051-35621-212-7-7-7)A=[051-321-2-7)-(5)B=[95605112-7-7)=[A)=0det([52-2-7])+5det([[-321-7])+1det([[[-3512]))+1det([[(A)=(5)-(5)-(5)d([(7))-(A)=(5)(A)+(7)=(5)(A)+(7)=(5(7)(A)=(5)(A) (A)b(B)b(t)-(4)-(B)-(B)t)-(4)-(4)A(A(4)A(4)A(A(4)=(A(4)A(4))A(A(B)=(A(B)=(4)A(B)=(A(B)=(A(4)=(B)()))(A(A(4)*(A(4)=(A(A(4)()))))(4)=(A(A(4)=(A(A()))))(A(A(A(A(A(4)))))))(A(A(A(A(4))))))(B)(A(A(A(B)))(B))(A(A(A(T)))))(A(A(A(A(A(T)=(A(4)))))))))(A(A(A(A(A(B)=(A(A(B)(T)(B))

:A)+1det([-321-7])+1det(A)+1det([-321-7])+1det([-3512])+(A)=0(5-(7)-2(2))-5(3-3(3)-2(7)-2(1))+(3(2)-(5)-(A)=0-(106)-(35-2)4)de(2-2)-(2-2)-(4)de(2-2)+1(6)-5)-(5)(A)=0det([522-7]))-([[-3211-7])-5det([[[-321-7])+(5det([[3512]))+(5det([(A)=0-95-11))d([(3))))d(A)*(A(4)&(B(B)-(T)A(4)A(4)A(4)A(4)(B)-(B)T)(B(B)t)T)=(A(3)(A(A(A(4)=(D)=(A(D)(D)(D))(A(A(D)(T)(A(T)))(A(T)))(A(A(A(T))(A(A(B)=(A(B)(T)(B)))))(D(T)(A(A(T)(T)(D)=(A(D)=(A(T))))))(A(A(A(A(A(A(A(D))))(D)))))))))(A(A(D(T)(T)(B(A(A(T))))))))(A(A(A(A(T))))(A(A(A(A(T)(D(T)))(B))(B))(A(A(A(A(T)))(B)(A(A(T)(B)))

Section outline

General

更多的决定因素特性