Any elementary row operation is equivalent to left multiplying by the corresponding elementary matrix ² Justification of LU Decomposition Algorithm Recall in
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[PDF] 25 Inverse Matrices - MIT Mathematics
It fails to have two pivots as required by Note 1 Elimination turns the second row of this matrix A into a zero row The Inverse of a Product AB
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M = I = M −1 M Inverse of a 2 × 2 Matrix Let M and N be the matrices: M = ( a b c d \ , N = ( d −b −c a \ Multiplying these matrices gives: MN = ( ad − bc 0
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1 fév 2012 · This is a requirement in order for matrix multiplication to be defined x A The notion of an inverse matrix only applies to square matrices
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Matrix Inverse: A –1 An n × n matrix A is said to have the inverse A–1 if the following two commutative matrix product relations are satisfied: A A–1 = I and
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an inverse matrix and how the inverse of a 2 × 2 matrix is calculated Preliminary example Suppose we calculate the product of the two matrices ( 4 3 1 1 ) and
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Any elementary row operation is equivalent to left multiplying by the corresponding elementary matrix ² Justification of LU Decomposition Algorithm Recall in
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24 jan 2013 · The matrix A can be expressed as a finite product of elementary matrices The matrix A has a left inverse (i e there exists a B such that BA = I)
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Diagonal matrices Inverse matrix Scalar multiplication: to multiply a matrix A by a scalar r, one The product of matrices A and B is defined if the number of
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Lecture6.InverseofMatrix
A~x= b:Inonedimensioncase,i.e.,Ais1£1;then
Ax=b canbeeasilysolvedas x= b A 1 A b=A ¡1 bprovidedthatA6=0:Denition7.1.AsquarematrixA
n£n issaidtobeinvertibleifthereexistsaunique matrixC n£n ofthesamesizesuchthatAC=CA=I
n C=A ¡1SupposenowA
n£n isinvertibleandC=A ¡1 isitsinversematrix.Thenthematrix equation A~x= b C=A ¡1 ;wehaveCA~x=C
b:Bydenition,CA=A
¡1 A=I n :Itleadsto I n ~x=C b; whichisthesameas ~x=A ¡1 b:(1) discussionissummarizedas A~x= b hasauniquesolution ~x=A ¡1 b: 1Example7.1(a)ShowAisinvertibleandA
¡1 =C;where A= 25¡3¡7
;C=¡7¡5
32(b)Solve 2x 1 +5x 2 =1
¡3x
1¡7x
2 =4: (c)Showthatthematrix B= 02 00 isNOTinvertible.Solution:(a)Directcalculationsleadto
AC= 25¡3¡7
¡7¡5
3210 01 =I 2 CA=
¡7¡5
3225
¡3¡7
10 01 =I 2Bydenition,
C=A ¡1 A~x= 1 4 ~x=A ¡1 1 4¡7¡5
321 4
¡27
11 (c)Bycalculation,wendthat B 2 02 00 02 00 00 00 BD=I 2 2B(BC)=BI
2 B 2 C=B; whichimplies 0=B; sinceB 2Theorem7.2A2£2matrix
A= ab cd det(A) def =ad¡bc6=0:Whenad¡bc6=0;theinverseis
A ¡1 1 ad¡bc d¡b¡ca
Example7.2.(a)FindA
¡1 if A= 25¡3¡6
(b)Solve2x+5y=1
¡3x¡6y=2
A ¡1 1 ad¡bc d¡b¡ca
1 3¡6¡5
32¡2¡5=3
12=3Wemayverifytheabovesolutionasfollows:
25¡3¡6
¡2¡5=3
12=3 10 01 b;where A= 25¡3¡6
b= 1 2Thesolutionis
~x=A ¡1 b=¡2¡5=3
12=3 1 2 16 3 7 3 3²PropertiesifInvertibleMatrix:
1.A ¡1 isalsoinvertibleand(A ¡1 ¡1 =A: 2.A T isalsoinvertibleand A T ¡1 =(A ¡1 T (AB) ¡1 =B ¡1 A ¡1Proof.(1)Bydenition,A
¡1 ifwecanndamatrixCsuchthat A ¡1 C=C A ¡1 =I:TheaboveisindeedtrueifC=A:
(2)Taketransposesofallthreesidesof(A ¡1 )A=A(A ¡1 )=I; A ¡1 A T A A ¡1 T =I T =)A T A ¡1 T A ¡1 T A T =I =)A T C=CA T =I;whereC= A ¡1 T istheinverseofA T (3)LetC=B ¡1 A ¡1 :SinceC(AB)=(CA)B=
B ¡1 A ¡1 A B= B ¡1 A ¡1 A B= B ¡1 I B= B ¡1 B=I (AB)C=A(BC)=A B B ¡1 A ¡1 =A BB ¡1 A ¡1 =A IA ¡1 =AA ¡1 =I: (AB) ¡1 =C=B ¡1 A ¡1 (4)SinceA~x=0hastheonlysolution
~x=A ¡1 0= 0; thushastobetheidentitymatrix. 4Atype(1)elementarymatrixE
1 instance, 2 4 100010 001 3 5 R 2 +¸R 1 !Rquotesdbs_dbs20.pdfusesText_26