推 ggssa2000:先END，推個。 11/02 00:42
推 frankho9:感謝分享^^ 11/02 19:18
餓死抬頭
又要考線代了
就照例來發個文騙騙文章數吧(誤)
第二章道通哥說是最簡單的一章
就好好來衝個高分吧
----------------------以上是廢話--------------------------
一.Determinants 行列式
矩陣求行列式值det(A)
(1)1x1 matrices
A = [a]
det(A) = a
(2)2x2 matrices
a11 a12
A = [ ]
a21 a22
det(A) = a11*a22 - a12*a21
(3)3x3 matrices
a11 a12 a13
A = [ a21 a22 a23 ]
a31 a32 a33
det(A) = a11(a22*a33-a23*a32)-a12(a21*a33-a23*a32+a13(a21*a32-a22*a31)
a22 a23
而我們令minor矩陣 M11 = [ ]
a32 a33
a21 a23
M12 = [ ]
a31 a33
a21 a22
M13 = [ ]
a31 a32
det(A)也等於a11*det(M11)-a12*det(M12)+a13*det(M13)
i+j
又令 cofactor Aij = (-1) *det(Mij)
=> 一個nxn的矩陣 A
a11 ,if n=1
det(A)= { }
a11*A11+a12*A12+···+a1n*A1n ,if n>1
-1
※若矩陣A的 det(A)≠0,A即為nonsingular,就會有反矩陣 A
證明:
U = Ek Ek-1 ··E1 A
det(U) = det(Ek)det(Ek-1)··det(E1)det(A)
≠0 ≠0 ≠0
若det(U)=0,則det(A)=0
得證
二.行列式的性質
T
(1)det(A ) = det(A)
(2)若矩陣A為三角矩陣,無論上三角(U)或下三角(L),det(A)=其對角線相乘
例:
1 2 3
A = [0 5 6] det(A) = 1*5*9 = 45
0 0 9
(3)矩陣中若任一列或任一行全為0,其行列式值=0
(4)矩陣中若有任兩行或任兩列值相同,其行列式值=0
例:
1 2 3
A = [4 5 6] det(A) = 0
4 5 6
(5)det(A+B) = det(A)+det(B)
(6)det(AB) = det(A)det(B)
(7)det(AB) = det(BA)
T T T T T
pf: det(AB) = det(AB ) = det(B A ) = det(B )det(A ) =det(B)det(A)
=det(BA)
三.Row Operation
det(EA) =det(E)det(A)
(1)Operation I (也就是前一章的type I,行列交換)
0 1
E =[ ] det(E) = -1
1 0
det(EA) = det(E)det(A) = (-1)det(A) = -det(A)
(2)Operation II (也就是前一章的type II,某行列值變α倍)
1 0 0
E =[0 1 0] det(E) = α
0 0 α
det(EA) = αdet(A)
(3)Operation III (也就是前一章的type III,行列倍數相加)
1 0 α
E =[0 1 0] det(E) = 1 (因為為三角型矩陣)
0 0 1
det(EA) = det(A)
四.
若A為一nxn的矩陣,Aij為其cofactor(就是前面一的Aij)
公式: ai1*Aj1 + ai2*Aj2 + ai3*Aj3+····+ain*Ajn
(1)若 i=j,原示 = a11*A11+a22*A22+···+ain*Ajn
= det(A)
(2)若 i≠j,則=0
證明: (道通哥說會考,原本說不會考的,突然改變主意....凸)
由(1)可得知
det(A) =ai1*Ai1+ai2*Ai2+···
(因為i=j)
=aj1*Aj1+aj2*Aj2+···
a11 a12 ··
···
矩陣 A =[ ai1 ··· ]
···
aj1···
而ai1要等於aj1才會有(1)的式子
亦即那兩列要為相等
而由determinants的性質可知
任兩行或兩列相等時其行列式值為0
得證
五.克拉馬公式 Cramer's Rule
(1) adjoint of a matrice
A11 A12 A13 T
adjA =[ A21 A22 A23 ]
A31 A32 A33
-1 1
A = adjA * ______
det(A)
證明: det(A) = A(adjA) = det(A)I
-1
A(adjA)(det(A)) = I
^^^^^^^^^^^^^^^^
-1
^^^^^^^^= A
得證
(2)求方程式解
AX=b
a11 a12 a13 b1
A =[a21 a22 a23] b=[b2]
a31 a32 a33 b3
b1 a12 a13 a11 b1 a13 a11 a12 b1
A1 = [ b2 a22 a23 ] , A2 = [ a21 b2 a23 ] , A3 = [ a21 a22 b2 ]
b3 a32 a33 a31 b3 a33 a31 a32 b3
* An就是把A的第n行換成b
det(A1) det(A2) det(A3)
則X1= _______ , X2= _______ ,X3= _______
det(A) det(A) det(A)
det(Ai)
=> Xi = _______ for i= 1,2,3,...,n
det(A)
證明: (我猜會考)
AX = b
-1 1
X = A b = _______ (adjA)b
det(A)
b1A1i+b2A2i+··+bnAni
Xi = ______________________
det(A)
det(Ai)
=_________
det(A)
得證
--
你在幹麻? 我在鋪梗阿
○ ╯ ╰ 更
/■\ ○/
/\ ╴/▔\
□□□□木更木更木更木更木更木 □□□□
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.204.166.225