Sis straipsnis apie matematine funkcija Apie rasto elementa ziurekite determinatyvas o apie kalbos dalį Kvadratines matricos determinantas algebrine suma visu galimu sandaugu gautu parenkant po viena dauginamajį is kiekvienos matricos eilutes taip kad dauginamieji priklausytu skirtingiems stulpeliams Determinantai svarbus integriniame ir diferenciniame skaiciavime geometrijoje kitose matematikos srityse Determinanto n n displaystyle n times n formule yra tokia d e t A A a 11 a 12 a 1 n a 21 a 22 a 2 n a n 1 a n 2 a n n i 1 n 1 p i a 1 k i 1 a 2 k i 2 a n k i n displaystyle det A A begin vmatrix a 11 amp a 12 amp ldots amp a 1n a 21 amp a 22 amp ldots amp a 2n ldots amp ldots amp ldots amp ldots a n1 amp a n2 amp ldots amp a nn end vmatrix sum i 1 n 1 p i cdot a 1k i1 a 2k i2 ldots a nk in kur A displaystyle A ir d e t A displaystyle det A determinanto zymejimas Antros eiles determinantas2 2 matrica A a b c d displaystyle A begin bmatrix a amp b c amp d end bmatrix turi determinanta det A a d b c displaystyle det A ad bc Determinantas taikomas spresti sistema su dviem nezinomaisiais a 11 x a 12 y c 1 a 21 x a 22 y c 2 displaystyle begin cases a 11 x a 12 y c 1 a 21 x a 22 y c 2 end cases Surandamas determinantas D a 11 a 12 a 21 a 22 a 11 a 22 a 12 a 21 displaystyle D begin vmatrix a 11 amp a 12 a 21 amp a 22 end vmatrix a 11 a 22 a 12 a 21 Jei determinantas nelygus nuliui tai sistema turi tik viena sprendinį x D x D displaystyle x frac D x D y D y D displaystyle y frac D y D kur D x c 1 a 12 c 2 a 22 displaystyle D x begin vmatrix c 1 amp a 12 c 2 amp a 22 end vmatrix D y a 11 c 1 a 21 c 2 displaystyle D y begin vmatrix a 11 amp c 1 a 21 amp c 2 end vmatrix Formules vadinamos Kramerio formulemis Jei D 0 bet D x displaystyle D x arba D y displaystyle D y nelygu 0 tai sistema sprendiniu neturi yra nesuderinta Jei D D x D y 0 displaystyle D D x D y 0 tai sistema turi be galo daug sprendiniu yra neapibrezta Pavyzdys kaip galima isspresti sistema surandant determinanta Sistema yra tokia x 2 y 8 3 x y 3 displaystyle begin cases x 2y 8 3x y 3 end cases Sistemos determinantas yra D 1 2 3 1 1 1 3 2 7 displaystyle D begin vmatrix 1 amp 2 3 amp 1 end vmatrix 1 cdot 1 3 cdot 2 7 Toliau į determinanto pirma stulpelį įstacius desines lygties puses randamas D x 8 2 3 1 8 6 14 displaystyle D x begin vmatrix 8 amp 2 3 amp 1 end vmatrix 8 6 14 Panasiai randamas D y 1 8 3 3 3 24 21 displaystyle D y begin vmatrix 1 amp 8 3 amp 3 end vmatrix 3 24 21 x D x D 14 7 2 displaystyle x D x D 14 7 2 y D y D 21 7 3 displaystyle y D y D 21 7 3 Determinantas 3 displaystyle times 3sudedami atimami d e t A A a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 displaystyle detA A begin vmatrix a 11 amp a 12 amp a 13 a 21 amp a 22 amp a 23 a 31 amp a 32 amp a 33 end vmatrix a 11 a 22 a 33 a 12 a 23 a 31 a 13 a 21 a 32 a 13 a 22 a 31 a 12 a 21 a 33 a 11 a 23 a 32 displaystyle a 11 a 22 a 33 a 12 a 23 a 31 a 13 a 21 a 32 a 13 a 22 a 31 a 12 a 21 a 33 a 11 a 23 a 32 Didesnems matricoms determinanto skaiciavimo formule yra kitokia Sistemu sprendimas taikant Kramerio formulesPagal Kramerio formule galima surasti sistemos a 11 x 1 a 12 x 2 a 13 x 3 c 1 a 21 x 1 a 22 x 2 a 23 x 3 c 2 a 31 x 1 a 32 x 2 a 33 x 3 c 3 displaystyle begin cases a 11 x 1 a 12 x 2 a 13 x 3 c 1 a 21 x 1 a 22 x 2 a 23 x 3 c 2 a 31 x 1 a 32 x 2 a 33 x 3 c 3 end cases sprendinius x 1 D 1 D displaystyle x 1 frac D 1 D x 2 D 2 D displaystyle x 2 frac D 2 D x 3 D 3 D displaystyle x 3 frac D 3 D kur D 1 c 1 a 12 a 13 c 2 a 22 a 23 c 3 a 32 a 33 displaystyle D 1 begin vmatrix c 1 amp a 12 amp a 13 c 2 amp a 22 amp a 23 c 3 amp a 32 amp a 33 end vmatrix D 2 a 11 c 1 a 13 a 21 c 2 a 23 a 31 c 3 a 33 displaystyle D 2 begin vmatrix a 11 amp c 1 amp a 13 a 21 amp c 2 amp a 23 a 31 amp c 3 amp a 33 end vmatrix D 3 a 11 a 12 c 1 a 21 a 22 c 2 a 31 a 32 c 3 displaystyle D 3 begin vmatrix a 11 amp a 12 amp c 1 a 21 amp a 22 amp c 2 a 31 amp a 32 amp c 3 end vmatrix Tokiu budu randami sistemos sprendiniai ir didesnems matricoms Remdamiesi Kramerio formulemis isspreskime tiesiniu lygciu sistema 2 x 1 3 x 2 x 3 7 3 x 1 4 x 2 6 x 3 3 3 x 1 x 2 3 x 3 5 displaystyle begin cases 2x 1 3x 2 x 3 7 3x 1 4x 2 6x 3 3 3x 1 x 2 3x 3 5 end cases D d e t A 2 1 1 3 4 6 1 0 1 3 1 1 3 4 6 0 0 1 1 3 3 3 1 3 4 15 displaystyle D detA begin vmatrix 2 amp 1 amp 1 3 amp 4 amp 6 1 amp 0 amp 1 end vmatrix begin vmatrix 3 amp 1 amp 1 3 amp 4 amp 6 0 amp 0 amp 1 end vmatrix 1 3 3 begin vmatrix 3 amp 1 3 amp 4 end vmatrix 15 Kur trecias stulpelis buvo padaugintas is 1 ir pridetas prie pirmo stulpelio trecias stulpelis nesikeicia D 1 0 1 1 5 4 6 1 0 1 0 0 1 5 10 6 1 1 1 1 1 1 3 5 10 1 1 15 displaystyle D 1 begin vmatrix 0 amp 1 amp 1 5 amp 4 amp 6 1 amp 0 amp 1 end vmatrix begin vmatrix 0 amp 0 amp 1 5 amp 10 amp 6 1 amp 1 amp 1 end vmatrix 1 cdot 1 1 3 begin vmatrix 5 amp 10 1 amp 1 end vmatrix 15 Kur trecias stulpelis buvo pridetas prie antro stulpelio D 2 2 0 1 3 5 6 1 1 1 0 0 1 15 5 6 3 1 1 1 1 1 3 15 5 3 1 30 displaystyle D 2 begin vmatrix 2 amp 0 amp 1 3 amp 5 amp 6 1 amp 1 amp 1 end vmatrix begin vmatrix 0 amp 0 amp 1 15 amp 5 amp 6 3 amp 1 amp 1 end vmatrix 1 cdot 1 1 3 begin vmatrix 15 amp 5 3 amp 1 end vmatrix 30 kur trecias stulpelis buvo padaugintas is 2 ir pridetas prie pirmojo stulpelio D 3 2 1 0 3 4 5 1 0 1 2 1 0 8 4 5 0 0 1 1 3 3 2 1 8 4 0 displaystyle D 3 begin vmatrix 2 amp 1 amp 0 3 amp 4 amp 5 1 amp 0 amp 1 end vmatrix begin vmatrix 2 amp 1 amp 0 8 amp 4 amp 5 0 amp 0 amp 1 end vmatrix 1 3 3 begin vmatrix 2 amp 1 8 amp 4 end vmatrix 0 kur trecias stuleplis buvo padaugintas is 1 ir pridetas prie pirmo stulpelio x 1 D 1 d e t A 15 15 1 x 2 D 2 d e t A 30 15 2 x 3 D 3 d e t A 0 15 0 displaystyle x 1 frac D 1 detA frac 15 15 1 qquad x 2 frac D 2 detA frac 30 15 2 qquad x 3 frac D 3 detA frac 0 15 0 Lygciu sprendimas atvirkstines matricos metoduDeterminanto radimas naudojant adjunkta d e t A 1 0 1 0 0 2 1 3 1 2 1 2 3 1 0 1 3 6 0 displaystyle detA begin vmatrix 1 amp 0 amp 1 0 amp 0 amp 2 1 amp 3 amp 1 end vmatrix 2 cdot 1 2 3 begin vmatrix 1 amp 0 1 amp 3 end vmatrix 6 not 0 kur 2 ir 3 virs 1 yra antra eilute ir trecias stulpelis A 11 1 1 1 0 2 3 1 6 A 12 1 1 2 0 2 1 1 2 displaystyle A 11 1 1 1 begin vmatrix 0 amp 2 3 amp 1 end vmatrix 6 qquad A 12 1 1 2 begin vmatrix 0 amp 2 1 amp 1 end vmatrix 2 A 13 1 1 3 0 0 1 3 0 A 21 1 2 1 0 1 3 1 3 displaystyle A 13 1 1 3 begin vmatrix 0 amp 0 1 amp 3 end vmatrix 0 qquad A 21 1 2 1 begin vmatrix 0 amp 1 3 amp 1 end vmatrix 3 A 22 1 2 2 1 1 1 1 2 A 23 1 2 3 1 0 1 3 3 displaystyle A 22 1 2 2 begin vmatrix 1 amp 1 1 amp 1 end vmatrix 2 qquad A 23 1 2 3 begin vmatrix 1 amp 0 1 amp 3 end vmatrix 3 A 31 1 3 1 0 1 0 2 0 A 32 1 3 2 1 1 0 2 2 displaystyle A 31 1 3 1 begin vmatrix 0 amp 1 0 amp 2 end vmatrix 0 qquad A 32 1 3 2 begin vmatrix 1 amp 1 0 amp 2 end vmatrix 2 A 33 1 3 3 1 0 0 0 0 displaystyle A 33 1 3 3 begin vmatrix 1 amp 0 0 amp 0 end vmatrix 0 Tiesiniu lygciu sistemos sprendimo metodas vadinamas atvirkstines matricos metodu arba matricu metodu A 1 1 d e t A A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 1 6 6 3 0 2 2 2 0 3 0 1 1 2 0 1 3 1 3 1 3 0 1 2 0 displaystyle A 1 frac 1 detA cdot begin vmatrix A 11 amp A 21 amp A 31 A 12 amp A 22 amp A 32 A 13 amp A 23 amp A 33 end vmatrix frac 1 6 cdot begin vmatrix 6 amp 3 amp 0 2 amp 2 amp 2 0 amp 3 amp 0 end vmatrix begin vmatrix 1 amp frac 1 2 amp 0 frac 1 3 amp frac 1 3 amp frac 1 3 0 amp frac 1 2 amp 0 end vmatrix Isspresime sistema 3 x 1 5 x 2 2 x 3 2 x 1 3 x 2 2 x 3 10 6 x 1 7 x 2 3 x 3 5 displaystyle begin cases 3x 1 5x 2 2x 3 2 x 1 3x 2 2x 3 10 6x 1 7x 2 3x 3 5 end cases matricu metodu A 3 5 2 1 3 2 6 7 3 B 2 10 5 X x 1 x 2 x 3 displaystyle A begin bmatrix 3 amp 5 amp 2 1 amp 3 amp 2 6 amp 7 amp 3 end bmatrix qquad B begin bmatrix 2 10 5 end bmatrix qquad X begin bmatrix x 1 x 2 x 3 end bmatrix X A 1 B A 1 1 d e t A A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 displaystyle X A 1 cdot B qquad A 1 frac 1 detA cdot begin bmatrix A 11 amp A 21 amp A 31 A 12 amp A 22 amp A 32 A 13 amp A 23 amp A 33 end bmatrix d e t A 3 5 2 1 3 2 6 7 3 0 14 8 1 3 2 0 25 15 1 2 1 14 8 25 15 2 5 7 4 5 3 10 0 displaystyle detA begin vmatrix 3 amp 5 amp 2 1 amp 3 amp 2 6 amp 7 amp 3 end vmatrix begin vmatrix 0 amp 14 amp 8 1 amp 3 amp 2 0 amp 25 amp 15 end vmatrix 1 2 1 begin vmatrix 14 amp 8 25 amp 15 end vmatrix 2 cdot 5 begin vmatrix 7 amp 4 5 amp 3 end vmatrix 10 not 0 Kur antra eilute padauginome is 3 ir pridejome prie pirmos eilutes ir antra eilute padauginome is 6 ir pridejome prie trecios eilutes A 11 1 1 1 3 2 7 3 5 A 21 1 2 1 5 2 7 3 1 A 31 5 2 3 2 4 displaystyle A 11 1 1 1 begin vmatrix 3 amp 2 7 amp 3 end vmatrix 5 qquad A 21 1 2 1 begin vmatrix 5 amp 2 7 amp 3 end vmatrix 1 qquad A 31 begin vmatrix 5 amp 2 3 amp 2 end vmatrix 4 A 12 1 2 6 3 15 A 22 1 2 2 3 2 6 3 3 A 32 3 2 1 2 8 displaystyle A 12 begin vmatrix 1 amp 2 6 amp 3 end vmatrix 15 qquad A 22 1 2 2 begin vmatrix 3 amp 2 6 amp 3 end vmatrix 3 qquad A 32 begin vmatrix 3 amp 2 1 amp 2 end vmatrix 8 A 13 1 3 6 7 25 A 23 1 2 3 3 5 6 7 9 A 33 3 5 1 3 14 displaystyle A 13 begin vmatrix 1 amp 3 6 amp 7 end vmatrix 25 qquad A 23 1 2 3 begin vmatrix 3 amp 5 6 amp 7 end vmatrix 9 qquad A 33 begin vmatrix 3 amp 5 1 amp 3 end vmatrix 14 A 1 1 10 5 1 4 15 3 8 25 9 14 displaystyle A 1 frac 1 10 begin bmatrix 5 amp 1 amp 4 15 amp 3 amp 8 25 amp 9 amp 14 end bmatrix X x 1 x 2 x 3 1 10 5 1 4 15 3 8 25 9 14 2 10 5 1 10 20 20 70 2 2 7 displaystyle X begin pmatrix x 1 x 2 x 3 end pmatrix frac 1 10 begin pmatrix 5 amp 1 amp 4 15 amp 3 amp 8 25 amp 9 amp 14 end pmatrix cdot begin pmatrix 2 10 5 end pmatrix frac 1 10 begin pmatrix 20 20 70 end pmatrix begin pmatrix 2 2 7 end pmatrix x 1 2 displaystyle x 1 2 x 2 2 displaystyle x 2 2 x 3 7 displaystyle x 3 7 Lygciu sistemos sprendimas Gauso metoduPavyzdziui turime lygciu sistema 3 x 1 2 x 2 4 x 3 8 2 x 1 7 x 2 5 x 3 26 x 1 3 x 2 8 x 3 25 displaystyle begin cases 3x 1 2x 2 4x 3 8 2x 1 7x 2 5x 3 26 x 1 3x 2 8x 3 25 end cases Isplestines matricos A pirmoje eiluteje parasome trecios eilutes koeficientus o pirma ir antra eilutes nustumiame zemyn A 1 3 8 25 3 2 4 8 2 7 5 26 displaystyle A begin bmatrix 1 amp 3 amp 8 amp 25 3 amp 2 amp 4 amp 8 2 amp 7 amp 5 amp 26 end bmatrix Sios pertvarkytos isplestines matricos A displaystyle A pirma eilute dauginame is 3 ir pridedame prie antros eilutes ir taip pat pirma eilute dauginame is 2 ir pridedame prie trecios eilutes ir tada gauname tokia isplestine matrica 1 3 8 25 0 7 20 67 0 13 21 76 displaystyle begin bmatrix 1 amp 3 amp 8 amp 25 0 amp 7 amp 20 amp 67 0 amp 13 amp 21 amp 76 end bmatrix Toliau matricos antra eilute dauginame is 2 ir pridedame prie trecios eilutes 1 3 8 25 0 7 20 67 0 1 19 58 displaystyle begin bmatrix 1 amp 3 amp 8 amp 25 0 amp 7 amp 20 amp 67 0 amp 1 amp 19 amp 58 end bmatrix Toliau trecia eilute dauginame is 7 ir pridedame prie antros eilutes ir gauname 1 3 8 25 0 0 113 339 0 1 19 58 displaystyle begin bmatrix 1 amp 3 amp 8 amp 25 0 amp 0 amp 113 amp 339 0 amp 1 amp 19 amp 58 end bmatrix Toliau antra ir trecia eilutes sukeiciame vietomis 1 3 8 25 0 1 19 58 0 0 113 339 displaystyle begin bmatrix 1 amp 3 amp 8 amp 25 0 amp 1 amp 19 amp 58 0 amp 0 amp 113 amp 339 end bmatrix Gauta matrica apibudina lygciu sistema x 1 3 x 2 8 x 3 25 x 2 19 x 3 58 113 x 3 339 displaystyle begin cases x 1 3x 2 8x 3 25 x 2 19x 3 58 113x 3 339 end cases Is paskutines lygties x 3 339 113 3 displaystyle x 3 frac 339 113 3 Is antros lygties surandame x 2 58 19 x 3 58 57 1 displaystyle x 2 58 19x 3 58 57 1 Is pirmos lygties randame x 1 25 3 x 2 8 x 3 25 3 24 2 displaystyle x 1 25 3x 2 8x 3 25 3 24 2 Lygciu sistema turi viena sprendinį 2 1 3 Ketvirtos eiles determinantasKetvirtos eiles determinantas gali buti paverstas trecios eiles determinantu pavyzdziui D 3 1 1 2 5 1 3 4 2 0 1 1 1 5 3 3 1 3 1 2 1 1 2 1 3 4 5 3 3 1 3 3 1 3 1 2 5 1 4 1 5 3 displaystyle D begin vmatrix 3 amp 1 amp 1 amp 2 5 amp 1 amp 3 amp 4 2 amp 0 amp 1 amp 1 1 amp 5 amp 3 amp 3 end vmatrix 1 3 1 cdot 2 cdot begin vmatrix 1 amp 1 amp 2 1 amp 3 amp 4 5 amp 3 amp 3 end vmatrix 1 3 3 cdot 1 cdot begin vmatrix 3 amp 1 amp 2 5 amp 1 amp 4 1 amp 5 amp 3 end vmatrix 1 3 4 1 3 1 1 5 1 3 1 5 3 2 16 40 48 40 displaystyle 1 3 4 cdot 1 cdot begin vmatrix 3 amp 1 amp 1 5 amp 1 amp 3 1 amp 5 amp 3 end vmatrix 2 cdot 16 40 48 40 Trecios eilutes antras stulpelis cia lygus 0 Ketvirtos eiles determinantui naudojamas Minoras M tai yra pries kiekviena sudetį yra isbraukiama eilute ir kiekvienas stulpelis kur yra skaicius toje eiluteje arba atvirksciai jei pasirenkamas pirma stulpelis Saltiniaideterminantas parenge Alfonsas Matuliauskas Visuotine lietuviu enciklopedija tikrinta 2024 02 03 Sis su matematika susijes straipsnis yra nebaigtas Jus galite prisideti prie Vikipedijos papildydami sį straipsnį