Chng III: TNH TON TRN MATLAB3.1 MngMng l mt tp hp s liu m ta mun tnh ton. to mng, t cc phn t ca mng vo gia 2 du ngoc vung3.1.1 Khai bo mng M = [a b c] hoc M = [a, b, c] VD: >> A = [3 6 4]hoc A = [3, 6, 4] % vector hng M = [ x; y; z]VD: >> B = [5; 1; 9] % vector ct Mng c t phn t.

Bin = gii hn u : gii hn cuihoc bin = gii hn u : bc chy : gii hn cuiVD: To 1 vect t chy t 0 n 0.6 vi bc chy tin l 0.1>> t = 0: 0.1:0.6 t = 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 Bin = linspace(gii hn u, gii hn cui, s phn t)VD: To 1 vect t chy t 0 n 10 vi 5 phn t gia>> t = linspace(0,10,5) t = 0 2.000 4.000 6.000 8.000 10.000 Mng c s lng ln cc phn t

3.1.2 Tnh ton dng s liu mngCc ton t nhn, chia, m phi thm du chm.*./.^VD: Tnh X = sin(A2)/(1+3A)>> X = sin (A.^2)./(1+3.*A)>> X = 0.0412-0.0522-0.02213.2 Ton t quan h quyt nh chng trnhNh hn: =Bng: == Khng bng (khc):~=

3.3 Ma trn3.3.1 nh ngha: Ma trn l mt mng nhiu chiu v theo nguyn tc:Bao quanh cc phn t ca ma trn bng du ngoc vung [aij].Cc phn t trong mt hng ca ma trn c cch nhau bi k t trng (space) hoc du phy (,).Kt thc mt hng trong ma trn bi du (;). Ni cch khc du (;) phn cch gia cc hng ca ma trn3.3.2 Nhp ma trnLit k trc tip:VD: >>A =[1 2 3; 4 5 6]>> B =[1 2 3; 4 5 6]

Nhp thng qua lnh input:>> input('Nhap gia tri cho ma tran C = ') >> Nhap gia tri cho ma tran C = [1 3 4;4 5 7;7 5 8] >> ans = 1 3 4 4 5 7 7 5 8 hin th li ma trn ta g tn ma trn sau enter. VD: >> AA = 1 2 3 4 5 6

3.3.3 Ma trn con v phn tTa c th ly mt phn t hoc ma trn con ca mt ma trn c.Tn ca ma trn (Ch s hng, ch s ct)A(i,j) s liu ca phn t hng i v ct j>> A(2,3)ans = 6>> B = C (2 : 3 , 1 : 3) B=4 5 7 7 5 8

3.3.4 Nhn 2 ma trnPhp nhn 2 ma trn l C = A*B.>> A = [6 -2; 10 3; 4 7];>> B = [9 8; -5 12];>> C = A*BC = 64 2475 1161 116Mun nhn 2 ma trn th s ct ca ma trn A phi bng s hng ca ma trn B

3.3.5 Hon v ma trnPhp chuyn i vct hng thnh vct ct gi l php hon v. Thc hin php chuyn v bng ton t du nhy n ( ). >> a = [1 2 3; 4 5 6; 7 8 9]a = 1 2 34 5 67 8 9>> b = ab = 1 4 72 5 83 6 9

3.3.6 Xa hng v ct ma trnMun xa hng hay ct ta gn gi tr rng (k hiu []) cho hng hoc ct ca ma trn.>> a = [1 2 3; 4 5 6; 7 8 9];>> a (2,:) = []a = 123789>> a (:,3) = []a = 1278

3.3.7 Cc ma trn c bitzeros(m,n)l ma trn m x n vi cc phn t bng 0ones(m,n)l ma trn m x n vi cc phn t bng 1eye(m,n)l ma trn m x n vi s 1 trn ng cho chnh>> zeros (2,3)ans = 000000>> B = [ones(2) zeros(2,3); zeros(3,2) 7*eye(3)]B = 11000 11000007000007000007

3.3.8 Cc hm ma trnexpm(A)tm hm m ca ma trn A, tc eAlogm(A)tm log(A)sqrtm(A)tm cn Adet(A)tnh nh thc A

3.4 ng dng cc php ton ma trn3.4.1 Gii phng trnh bc cao: anxn + an-1xn-1+...+a0 = 0Bc1: Lp 1 ma trn hng c cc phn t l cc h s t an n a0 gim dn theo bc ca phng trnh (Nu h s no khng c ghi 0)Bc 2: Dng lnh roots gii ma trn va to c.VD: gii phng trnh sau x5 - 2x4 + 5x2 - 1 = 0>> y = [ 1 -2 050-1]; >> kq = roots(y) kq =1.5862 + 1.1870i1.5862 - 1.1870i-1.1606 -0.4744 0.4627

3.4.2 Bit nghim tm li phng trnh S dng lnh poly. >> r = [1, 3 + 5i, 3 - 5i];>> poly(r)ans =1 -7 -40 34

3.4 ng dng cc php ton ma trn

3.4 ng dng cc php ton ma trn

Nh vy vic gii h PT tuyn tnh thc cht l thc hin php ton v ma trn.V vy, nghim ca h phng trnh l: u = A\bVic gii trong Matlab gm 3 dng sau:>> b = [8; -6; 4];>> A = [2 4 -3; -2 3 2; -2 4 1];>> u = A\bCh : du \ l du chia tri dng gii h phng trnh khi h ch c mt nghim