DSC Chuong 1

CHNG 1MA TRN V H PHNG TRNH I S TUYN TNH-----

Chng 1. Ma trn H PT STT

3) Nhn hai ma trn:


V d:

Tnh cht ca tch cc ma trn:

nh l 4:


Chng minh (1) K hiu: Dmxp=AmxnBnxp, Emxq=(AB)C=DmxpCpxq Fnxq=BnxpCpxq, Gmxq=A(BC)=AmxnFnxq Ta cn cm: E=GTnh : Dmxp?Cc phn t hng 1 ca D:Phn t d11? Chng 1. Ma trn H PT STT

Cc phn t hng 1 ca D:Tnh Emxq?:Tnh e11: Chng 1. Ma trn H PT STT

Tnh eij:Vy E=G (pcm) Chng 1. Ma trn H PT STT

Ch : 1) Tn ti AB v BA, trong trng hp tng qut AB khc BA (php nhn 2 ma trn khng c tnh giao hon). Ngc li, nu AB=BA th ta ni A v B giao hon vi nhau 2) Gi s: A khc khng v B khc khng, th c th AB=0. Chng 1. Ma trn H PT STT

V d: V d: Do khng nh AB=0 th A=0 hay B=0 l sai.

nh l 5: Cho Amxn, Bnxp th Chng 1. Ma trn H PT STT

Chng minh:

L tha ma trn:

nh ngha:Cho A l ma trn vung cp n. Ta gi ly tha bc k (k: s nguyn) ca A l mt ma trn cp n (k hiu Ak) c xc nh mt cch quy np nh sauMa trn ly linh:

nh ngha:Cho A l ma trn vung cp n tha iu kin Ak=0 vi mt s nguyn k no th A gi l mt ma trn ly linh.V d: Chng 1. Ma trn H PT STT

Tnh cht:

Cho A l ma trn vung cp n, r v sl hai s nguyn Chng 1. Ma trn H PT STT


Php bin i s cp trn dng (ct) ca ma trn


Ma trn dng bc thang chnh tc:Nu mt hng c mt s khc khng th s khc khng bn tri nht bng 1, c gi l phn t chnh.

Nhng hng gm ton nhng phn t khng nm di cng.

Nu hai hng k nhau c phn t chnh th phn t chnh ca hng trn nm bn tri phn t chnh hng di.

Mi ct c phn t chnh th cc phn t khc u bng khng.

Ma trn dng bc thang:Ma trn bc thang c cc dng khc 0 nm bn trn cc dng 0.

Trn hai dng khc khng, phn t khc khng u tin ca dng di nm bn phi phn t khc khng u tin ca dng trn.


Dng 0 l dng gm tt c cc phn t bng 0.

5. Ma trn vung kh nghch:Cho A l ma trn vung cp n khc khng,ta ni A kh nghch khi tn ti ma trn Bcng cp vi A sao cho: AB=BA=In. Khi ta ni B l ma trn nghch o ca A, khiu: B=A-1. Chng 1. Ma trn H PT STT

nh ngha:

Nu A khng kh nghch, ta ni A suy bin.

Nu B l ma trn nghch o ca A th A cng l ma trn nghch o ca B.


Nu A c mt dng bng 0 (hay mt ct bng 0) th A suy bin.

Tnh cht:

Ma trn nghch o ca A (nu c) l duy nht.

Nu A kh nghch th AT, A (0), A-1 cng kh nghch v hn na:

Nu A v B cng kh nghch th AB cng kh nghch v (AB)-1=B-1A-1.

Nu A1, A2,,An cng kh nghch th tch ca chng cng kh nghch v (A1A2An)-1=An-1An-1-1A1-1.


Tm ma trn nghch o bng php bin i s cp trn dng:

Cho ma trn vung A cp n:Bc 2: Dng cc php bin i s cp trn dng a [A|I] v dng [A|B]. C 2 trng hp:Bc 1: Lp ma trn c dng [An|In].MT A c mt dng (hoc mt ct) bng 0. Ta dng li v kt lun A suy bin.

MT A=In. Ta dng li v kt lun A kh nghch, v ma trn nghch o A-1=B.


5. H phng trnh i s tuyn tnh:Mt h PT STT trn R l mt h gm m phng trnh bc nht (n n) c dng tng qut:nh ngha:

trong cc aij (gi l cc h s) v cc bi (cc h s t do) l cc phn t cho trc, cc xj l cc n cn tm.(*)Ta ni (c1,c2,,cn) l nghim ca h (*) nu khi thay x1=c1, x2=c2,xn=cn vo (*) th tt c cc ng thc trong (*) u tha.

Nu cc h s t do bi=0 th h tr thnh h PT STT thun nht.


H PT STT thun nht c t nht mt nghim l (x1,x2,,xn)=(0,0,,0): gi l nghim tm thng.

nh l:

i vi mt h PT STT th ch c mt trong 3 trng hp nghim:C nghim duy nht,

C v s nghim,

V nghim.

H qu:

H PT STT thun nht ch c:Nghim tm thng,

V s nghim.

H (*) c vit li di dng ma trn nh sau:


Ma trn h s.Ma trn n s.Ma trn hng s.

K hiu:


Ma trn h s m rng ca h (*).

H 2 PT STT (c cng s n) c gi l tng ng nhau nu n c cng tp nghim.nh ngha:


Cho hai h gm m phng trnh tuyn tnh n n sao cho ma trn h s m rng ca hai h ln lt l v . Khi nu th hai h trn tng ng nhau.nh l:

i vi h PT thun nht c ct cc h s hng bng 0, nn khi gii ta khng cn lp ma trn h s m rng m ch cn ly ma trn h s bin i.


Hai ma trn tng ng nhau khi v ch khi tn ti mt s hu hn cc php BSC trn dng bin ma trn ny thnh ma trn cn li.

Lu :

T h PT STT ban u ta s dng cc php BSC trn dng ty i vi ma trn h s m rng ca h ny a n v dng mt h PT STT n gin hn.


Phng php Gauss Jordan gii h PT STT:

Gm 3 bc:Bc 1:Thit lp ma trn h s m rng: Bc 2:S dng cc php BSC trn dng rt gn ma trn h s m rng v dng bc thang chnh tc.Bc 3:Bin lun nghim.


VD 2:Gii h PT STT sau: Bc 1:Thit lp ma trn h s m rng: Bc 2:Tin hnh thut ton Gauss - Jordan:


Bc 3: H c nghim duy nht

Phng php Gauss gii h PT STT:

Gm 3 bc:Bc 1:Thit lp ma trn h s m rng. Bc 2:S dng cc php BSC trn dng rt gn ma trn h s m rng v dng bc thang.Bc 3:Bin lun nghim. Chng 1. Ma trn H PT STT


VD 2:Gii h PT STT sau: Bc 1:Thit lp ma trn h s m rng: Bc 2:Tin hnh thut ton Gauss:


Bc 3: H c nghim duy nht

Nhn xt:Trong qu trnh bin i nu: Chng 1. Ma trn H PT STT

C cc dng bng 0 th ta loi dng i.

C 2 dng t l vi nhau th ta loi mt trong 2 dng i.

Nu mt dng c dng [0 0 0|b] vi b khc khng th h PT v nghim.

Gii h PT STT bng ma trn nghch o:


Xt h PT STT gm n phng trnh v n n s c dng AX=B.nh l:

Nu A khng suy bin th h phng trnh c mt nghim duy nhtX=A-1B

Chng 1. Ma trn nh thc H pttt

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