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Determinanty

Aplikovana matematika I

Dana Rıhova

Mendelu Brno

c©Dana Rıhova (Mendelu Brno) Determinanty 1 / 38

Obsah

1 DeterminantyDefinice determinantuSarrusovo a krızove pravidloVlastnosti determinantuLaplaceuv rozvojVypocet determinantu radu n ≥ 4

2 Inverznı matice3 Maticove rovnice

Pierre Simon de Laplace


Permutace

Definice (permutace)

Necht’ jsou dana cısla 1, 2, 3, . . . , n. Kazda usporadana n-tice utvorena z techtoprvku se nazyva permutace.

Pocet vsech ruznych permutacı skupiny n prvku je n!.

Prıklad (permutace)

Permutace trı prvku 1, 2, 3 jsou trojice

(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1).

Jejich pocet je 3! = 3 · 2 · 1 = 6.


Permutace






(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1).

Jejich pocet je 3! = 3 · 2 · 1 = 6.


Permutace






(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1).

Jejich pocet je 3! = 3 · 2 · 1 = 6.


Definice (inverze)

Necht’ (k1, k2, . . . , kn) je nejaka permutace cısel 1, 2, 3, . . . , n. Rıkame, ze dvojice(ki, kj), kde i < j, tvorı v teto permutaci inverzi, je-li ki > kj .

Prıklad (pocet inverzı)

1 V permutaci (2, 4, 1, 3) cısel 1, 2, 3, 4 tvorı kazda z dvojic (2, 1), (4, 1), (4, 3)inverzi. Permutace ma 3 inverze.

2 Permutace (4, 2, 1, 3) cısel 1, 2, 3, 4 ma celkem 4 inverze: (4, 2), (4, 1), (4, 3),(2, 1).


Definice (inverze)






Definice (inverze)






Determinant

Definice (determinant)

Necht’ A je ctvercova matice radu n. Determinantem matice A nazyvame realnecıslo

detA =∑

(−1)pa1k1a2k2

. . . ankn,

kde scıtame pres vsechny permutace (k1, k2, . . . , kn) sloupcovych indexu. Cıslo pudava pocet inverzı prıslusne permutace. Znacıme

detA = |A| =

∣∣∣∣∣∣∣∣∣a11 a12 · · · a1na21 a22 · · · a2n...

.... . .

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣ .


Poznamka (determinant)

Determinant matice je definovan pouze pro ctvercovou matici.

Determinant matice je soucet soucinu prvku matice vybranych tak, zez kazdeho radku a z kazdeho sloupce je vybran prave jeden prvek.Prvky prıslusneho soucinu jsou usporadany tak, aby na prvnım mıste bylprvek z 1. radku, na druhem mıste z 2. radku atd. Souciny jsou opatrenyznamenkem podle permutace sloupcovych indexu.

Poznamka (hlavnı a vedlejsı diagonala)

Prvky a11, a22, · · · , ann tvorı hlavnı diagonalu,prvky an1, an−1,2, · · · , a1n tvorı vedlejsı diagonalu.














Vypocet determinantu matic radu 1, 21 Determinant matice radu 1:

detA = |a11| = a11

2 Determinant matice radu 2 (krızove pravidlo):

detA =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinant se rovna soucinu prvku hlavnı diagonaly minus soucin prvkuvedlejsı diagonaly.

schema pro zapamatovanı:

+−a21

a11

a22

a12= a11a22 − a12a21



detA = |a11| = a11


detA =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



+−a21

a11

a22

a12= a11a22 − a12a21



detA = |a11| = a11


detA =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



+−a21

a11

a22

a12= a11a22 − a12a21



detA = |a11| = a11


detA =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



+−a21

a11

a22

a12= a11a22

− a12a21



detA = |a11| = a11


detA =

∣∣∣∣ a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



+−a21

a11

a22

a12= a11a22 − a12a21


Vypocet determinantu matic radu 33 Determinant matice radu 3 (Sarrusovo pravidlo):

detA =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ = a11a22a33+a13a21a32+a12a23a31−a13a22a31−a11a23a32−a12a21a33


+−

+−

+−

a31

a21

a32

a22

a11 a12

a33

a23

a13

a11 a12 a13

a21 a22 a23

= a11a22a33 + a21a32a13 + a31a12a23− a13a22a31 − a23a32a11 − a33a12a21

Prvnı dva radky opıseme pod determinant, setrojıme souciny podlenaznacenych spojnic a opatrıme je uvedenymi znamenky.



detA =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33



+−

+−

+−

a31

a21

a32

a22

a11 a12

a33

a23

a13

a11 a12 a13

a21 a22 a23

= a11a22a33 + a21a32a13 + a31a12a23

− a13a22a31 − a23a32a11 − a33a12a21




detA =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33



+−

+−

+−

a31

a21

a32

a22

a11 a12

a33

a23

a13

a11 a12 a13

a21 a22 a23





detA =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33



+−

+−

+−

a31

a21

a32

a22

a11 a12

a33

a23

a13

a11 a12 a13

a21 a22 a23




Prıklad (krızove a Sarrusovo pravidlo)1 krızove pravidlo:∣∣∣∣ 2 −1

3 1

∣∣∣∣

= 2 · 1− (−1) · 3 = 2 + 3 = 5

2 Sarrusovo pravidlo:∣∣∣∣∣∣−2 3 −11 2 −10 3 4

∣∣∣∣∣∣ =

(−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37

Poznamka (vypocet determinantu matic radu n ≥ 4)

Pro vypocet determinantu matic ctvrteho a vyssıch radu neexistuje zadne obdobnepravidlo k Sarrusovu a krızovemu pravidlu.



3 1

∣∣∣∣ = 2 · 1− (−1) · 3

= 2 + 3 = 5


∣∣∣∣∣∣ =

(−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ =

(−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ =

(−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ =

(−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5

2 Sarrusovo pravidlo:∣∣∣∣∣∣− 2 3 −1

1 2 −10 3 4

∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ = (−2) · 2 · 4 + 1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 − 11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1) + 0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 − 1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5

2 Sarrusovo pravidlo:∣∣∣∣∣∣−2 3 − 11 2 −10 3 4

∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)− (−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5

2 Sarrusovo pravidlo:∣∣∣∣∣∣−2 3 −11 2 − 10 3 4

∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0− (−1) · 3 · (−2)−4 · 3 · 1

− 2 3 −11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)− 4 · 3 · 1

−2 3 −11 2 −1

= −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37





3 1

∣∣∣∣ = 2 · 1− (−1) · 3 = 2 + 3 = 5


∣∣∣∣∣∣ = (−2) · 2 · 4+1 · 3 · (−1)+0 · 3 · (−1)−(−1) · 2 · 0−(−1) · 3 · (−2)−4 · 3 · 1

−2 3 −11 2 −1 = −16− 3 + 0− 0− 6− 12 = −19− 18 = −37




Vlastnosti determinantu

Veta (determinant transponovane matice)

Pro libovolnou ctvercovou matici platı

detA = detAT .

Poznamka (determinant transponovane matice)

Transponovanım matice se hodnota determinantu nezmenı, muzeme v nemtedy zamenit radky za sloupce a naopak.

Vsechny vlastnosti determinantu, vyslovene pro radky, platı take pro sloupce.

Prıklad (determinant transponovane matice)∣∣∣∣ 2 −13 0

∣∣∣∣ = ∣∣∣∣ 2 3−1 0

∣∣∣∣ = 3,

∣∣∣∣∣∣5 1 0−1 4 32 0 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣5 −1 21 4 00 3 −1

∣∣∣∣∣∣ = −15





detA = detAT .





∣∣∣∣ = ∣∣∣∣ 2 3−1 0

∣∣∣∣ = 3,

∣∣∣∣∣∣5 1 0−1 4 32 0 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣5 −1 21 4 00 3 −1

∣∣∣∣∣∣ = −15





detA = detAT .





∣∣∣∣ = ∣∣∣∣ 2 3−1 0

∣∣∣∣ = 3,

∣∣∣∣∣∣5 1 0−1 4 32 0 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣5 −1 21 4 00 3 −1

∣∣∣∣∣∣ = −15





detA = detAT .





∣∣∣∣ = ∣∣∣∣ 2 3−1 0

∣∣∣∣ = 3,

∣∣∣∣∣∣5 1 0−1 4 32 0 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣5 −1 21 4 00 3 −1

∣∣∣∣∣∣ = −15c©Dana Rıhova (Mendelu Brno) Determinanty 10 / 38


Veta (vlastnosti determinantu nemenıcı jeho hodnotu)1 Hodnota determinantu matice se nezmenı, pricteme-li k jednomu radku

(sloupci) nasobek jineho radku (sloupce).

Veta (vlastnosti determinantu menıcı jeho hodnotu)

2 Vymenıme-li v determinantu mezi sebou dva radky (sloupce), determinantzmenı znamenko.

3 Vynasobıme-li jeden radek (sloupec) nenulovym cıslem k, zvetsı se hodnotadeterminantu k-krat.

Poznamka (vlastnosti determinantu)

Uvedene vlastnosti se pouzıvajı pri upravach determinantu vyssıch radu, kdypomocı nich vytvorıme v nekterem radku, resp. sloupci co nejvıce nul.




















Prıklad (vlastnosti determinantu)

1

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 4 33 4

∣∣∣∣ = 7,

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 0 7−1 1

∣∣∣∣ = 7

prictenı 1. radku k 2. radku prictenı 4-nasobku 2. radku k 1. radku

2

∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ 4 12 −3

∣∣∣∣ , ∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ −3 21 4

∣∣∣∣14 = − (−14) 14 = − (−14)zamena radku zamena sloupcu

3

∣∣∣∣ −2 13 2

∣∣∣∣ = 1

4

∣∣∣∣ −8 43 2

∣∣∣∣ , ∣∣∣∣ −2 13 2

∣∣∣∣ = 1

3

∣∣∣∣ −2 33 6

∣∣∣∣−7 =

1

4· (−28) −7 =

1

3· (−21)

vynasobenı 1. radku cıslem 4 vynasobenı 2. sloupce cıslem 3



1

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 4 33 4

∣∣∣∣ = 7,

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 0 7−1 1

∣∣∣∣ = 7


2

∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ 4 12 −3

∣∣∣∣ , ∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ −3 21 4


3

∣∣∣∣ −2 13 2

∣∣∣∣ = 1

4

∣∣∣∣ −8 43 2

∣∣∣∣ , ∣∣∣∣ −2 13 2

∣∣∣∣ = 1

3

∣∣∣∣ −2 33 6

∣∣∣∣−7 =

1

4· (−28) −7 =

1

3· (−21)




1

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 4 33 4

∣∣∣∣ = 7,

∣∣∣∣ 4 3−1 1

∣∣∣∣ = ∣∣∣∣ 0 7−1 1

∣∣∣∣ = 7


2

∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ 4 12 −3

∣∣∣∣ , ∣∣∣∣ 2 −34 1

∣∣∣∣ = − ∣∣∣∣ −3 21 4


3

∣∣∣∣ −2 13 2

∣∣∣∣ = 1

4

∣∣∣∣ −8 43 2

∣∣∣∣ , ∣∣∣∣ −2 13 2

∣∣∣∣ = 1

3

∣∣∣∣ −2 33 6

∣∣∣∣−7 =

1

4· (−28) −7 =

1

3· (−21)



Poznamka (vytykanı pred determinant)

Z vlastnosti 3 plyne, ze jsou-li prvky nektereho radku (sloupce) nasobkemnejakeho cısla, muzeme toto cıslo vytknout pred determinant.

Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣

= 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣

= 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣

= 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣

= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣

= 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)]

= 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44]

= 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300




Prıklad∣∣∣∣∣∣30 15 103 2 624 10 8

∣∣∣∣∣∣ = 3

∣∣∣∣∣∣10 15 101 2 68 10 8

∣∣∣∣∣∣ = 3 · 5

∣∣∣∣∣∣2 3 21 2 68 10 8

∣∣∣∣∣∣ = 3 · 5 · 2

∣∣∣∣∣∣2 3 21 2 64 5 4

∣∣∣∣∣∣= 3 · 5 · 2 · 2

∣∣∣∣∣∣2 3 11 2 34 5 2

∣∣∣∣∣∣ = 60[8 + 36 + 5− (8 + 30 + 6)] = 60[49− 44] = 60 · 5

= 300



Veta (nulovost determinantu)

Determinant matice se rovna nule, je-li jeden z jeho radku (sloupcu) linearnıkombinacı ostatnıch radku (sloupcu).

Determinant je roven nule, jsou-li jeho radky nebo sloupce linearne zavisle.

Poznamka (nulovost determinantu)

Determinant je nulovy zejmena, jestlize

nektery jeho radek (sloupec) je nulovy,

nektery radek (sloupec) je nasobkem jineho radku (sloupce),

dva radky (sloupce) jsou stejne.






















Prıklad (nulovost determinantu)

∣∣∣∣∣∣2 −3 −1−1 0 23 0 −6

∣∣∣∣∣∣ = 0, protoze 3. radek je (-3)-nasobkem 2. radku

∣∣∣∣ −1 −60 0

∣∣∣∣ = 0, protoze 2. radek je nulovy

∣∣∣∣∣∣−2 1 13 −1 −10 2 2

∣∣∣∣∣∣ = 0, protoze 2. a 3. sloupec jsou stejne



∣∣∣∣∣∣2 −3 −1−1 0 23 0 −6


∣∣∣∣ −1 −60 0


∣∣∣∣∣∣−2 1 13 −1 −10 2 2




∣∣∣∣∣∣2 −3 −1−1 0 23 0 −6


∣∣∣∣ −1 −60 0


∣∣∣∣∣∣−2 1 13 −1 −10 2 2



Veta (vypocet determinantu - soucin prvku na hlavnı diagonale)

Determinant matice, ktera ma pod hlavnı diagonalou same nuly, je roven soucinuprvku na jejı hlavnı diagonale.

Prıklad (vypocet determinantu - soucin prvku na hlavnı diagonale)∣∣∣∣∣∣∣∣1 2 3 4−1 0 3 4−1 −2 0 4−1 −2 −3 4

∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−−−−−+

=

∣∣∣∣∣∣∣∣1 2 3 40 2 6 80 0 3 80 0 0 8

∣∣∣∣∣∣∣∣ = 1 · 2 · 3 · 8 = 48

1. radek jsme postupne pricetli k 2., 3. a 4. radku





∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−−−−−+

=

∣∣∣∣∣∣∣∣1 2 3 40 2 6 80 0 3 80 0 0 8

∣∣∣∣∣∣∣∣ = 1 · 2 · 3 · 8 = 48






∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−−−−−+

=

∣∣∣∣∣∣∣∣1 2 3 40 2 6 80 0 3 80 0 0 8

∣∣∣∣∣∣∣∣

= 1 · 2 · 3 · 8 = 48






∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−−−−−+

=

∣∣∣∣∣∣∣∣1 2 3 40 2 6 80 0 3 80 0 0 8

∣∣∣∣∣∣∣∣ = 1 · 2 · 3 · 8 = 48



Vypocet determinantu matic vyssıch radu

Vypocet determinantu matic ctvrteho a vyssıch radu prevadıme na vypocetdeterminantu nizsıch radu pomocı Laplaceova rozvoje.

Definice (subdeterminant, algebraicky doplnek)

Necht’ A je ctvercova matice radu n. Vynechame-li v determinantu matice A i-tyradek a j-ty sloupec, dostaneme determinant Mij matice radu n− 1, kterynazyvame subdeterminant (minor) prıslusny prvku aij .

Algebraickym doplnkem Aij prvku aij v determinantu matice A nazyvame cıslo

Aij = (−1)i+jMij.

Indexy i, j u subdeterminantu Mij vyjadrujı, ze jsme v determinantuvynechali i-ty radek a j-ty sloupec.







Aij = (−1)i+jMij.








Aij = (−1)i+jMij.








Aij = (−1)i+jMij.



Prıklad (subdeterminant, algebraicky doplnek)

Mejme dan determinant detA =

∣∣∣∣∣∣2 3 2−1 1 10 5 4

∣∣∣∣∣∣ .

subdeterminant a algebraicky doplnek prvku a32 jsou

A =

∣∣∣∣∣∣2 3 2−1 1 1

0 5 4

∣∣∣∣∣∣ ⇒ M32 =

∣∣∣∣ 2 2−1 1

∣∣∣∣ = 2 + 2 = 4,

A32 = (−1)3+2

∣∣∣∣ 2 2−1 1

∣∣∣∣ = (−1) · 4 = −4


A =

∣∣∣∣∣∣2 3 2

-1 1 10 5 4

∣∣∣∣∣∣ ⇒ M21 =

∣∣∣∣ 3 25 4

∣∣∣∣ = 12− 10 = 2,

A21 = (−1)2+1

∣∣∣∣ 3 25 4

∣∣∣∣ = (−1) · 2 = −2




∣∣∣∣∣∣2 3 2−1 1 10 5 4

∣∣∣∣∣∣ .subdeterminant a algebraicky doplnek prvku a32 jsou

A =

∣∣∣∣∣∣2 3 2−1 1 1

0 5 4

∣∣∣∣∣∣ ⇒ M32 =

∣∣∣∣ 2 2−1 1

∣∣∣∣ = 2 + 2 = 4,

A32 = (−1)3+2

∣∣∣∣ 2 2−1 1

∣∣∣∣ = (−1) · 4 = −4


A =

∣∣∣∣∣∣2 3 2

-1 1 10 5 4

∣∣∣∣∣∣ ⇒ M21 =

∣∣∣∣ 3 25 4

∣∣∣∣ = 12− 10 = 2,

A21 = (−1)2+1

∣∣∣∣ 3 25 4

∣∣∣∣ = (−1) · 2 = −2




∣∣∣∣∣∣2 3 2−1 1 10 5 4


A =

∣∣∣∣∣∣2 3 2−1 1 1

0 5 4

∣∣∣∣∣∣ ⇒ M32 =

∣∣∣∣ 2 2−1 1

∣∣∣∣ = 2 + 2 = 4,

A32 = (−1)3+2

∣∣∣∣ 2 2−1 1

∣∣∣∣ = (−1) · 4 = −4


A =

∣∣∣∣∣∣2 3 2

-1 1 10 5 4

∣∣∣∣∣∣ ⇒ M21 =

∣∣∣∣ 3 25 4

∣∣∣∣ = 12− 10 = 2,

A21 = (−1)2+1

∣∣∣∣ 3 25 4

∣∣∣∣ = (−1) · 2 = −2




∣∣∣∣∣∣2 3 2−1 1 10 5 4


A =

∣∣∣∣∣∣2 3 2−1 1 1

0 5 4

∣∣∣∣∣∣ ⇒ M32 =

∣∣∣∣ 2 2−1 1

∣∣∣∣ = 2 + 2 = 4,

A32 = (−1)3+2

∣∣∣∣ 2 2−1 1

∣∣∣∣ = (−1) · 4 = −4


A =

∣∣∣∣∣∣2 3 2

-1 1 10 5 4

∣∣∣∣∣∣ ⇒ M21 =

∣∣∣∣ 3 25 4

∣∣∣∣ = 12− 10 = 2,

A21 = (−1)2+1

∣∣∣∣ 3 25 4

∣∣∣∣ = (−1) · 2 = −2




∣∣∣∣∣∣2 3 2−1 1 10 5 4


A =

∣∣∣∣∣∣2 3 2−1 1 1

0 5 4

∣∣∣∣∣∣ ⇒ M32 =

∣∣∣∣ 2 2−1 1

∣∣∣∣ = 2 + 2 = 4,

A32 = (−1)3+2

∣∣∣∣ 2 2−1 1

∣∣∣∣ = (−1) · 4 = −4


A =

∣∣∣∣∣∣2 3 2

-1 1 10 5 4

∣∣∣∣∣∣ ⇒ M21 =

∣∣∣∣ 3 25 4

∣∣∣∣ = 12− 10 = 2,

A21 = (−1)2+1

∣∣∣∣ 3 25 4

∣∣∣∣ = (−1) · 2 = −2


Veta (Laplaceuv rozvoj)

Pro kazdy determinant matice A radu n ≥ 2 platı

detA = ai1Ai1 + ai2Ai2 + · · ·+ ainAin pro i = 1, 2, . . . , n,

(Laplaceuv rozvoj determinantu podle i-teho radku),

detA = a1jA1j + a2jA2j + · · ·+ anjAnj pro j = 1, 2, . . . , n,

(Laplaceuv rozvoj determinantu podle j-teho sloupce)

kde Aij je algebraicky doplnek prvku aij .

Poznamka (Laplaceuv rozvoj)

Determinant se rovna souctu vsech soucinu prvku libovolneho radku(sloupce) a jejich algebraickych doplnku.

Tyto vzorce se pouzıvajı pro vypocet determinantu matic radu 4 a vyssıho.

Radek nebo sloupec, podle ktereho provadıme rozvoj, je vhodne volit tak, abyobsahoval co nejvıce nulovych prvku.


















































Prıklad (Laplaceuv rozvoj)

∣∣∣∣∣∣∣∣−1 1 2 21 0 −1 1−1 3 2 0−1 0 5 3

∣∣∣∣∣∣∣∣ =1 · (−1)1+2

∣∣∣∣∣∣1 −1 1−1 2 0−1 5 3

∣∣∣∣∣∣+ 0 · (−1)2+2

∣∣∣∣∣∣−1 2 2−1 2 0−1 5 3

∣∣∣∣∣∣+3 · (−1)3+2

∣∣∣∣∣∣−1 2 21 −1 1−1 5 3

∣∣∣∣∣∣+ 0 · (−1)4+2

∣∣∣∣∣∣−1 2 21 −1 1−1 2 0

∣∣∣∣∣∣

= (−1)[6 + 0− 5− (−2 + 0 + 3)]− 3[3− 2 + 10− (2− 5 + 6)]

= (−1)[1− 1]− 3[11− 3] = −24

Provedli jsme rozvoj podle 2. sloupce, protoze obsahoval nejvıce nul. Vznikledeterminanty tretıho radu jsme vypocetli pomocı Sarrusova pravidla. Pocıtali jsmepouze ty, ktere nebyly nasobeny nulovym prvkem.



∣∣∣∣∣∣∣∣−1 1 2 21 0 −1 1−1 3 2 0−1 0 5 3

∣∣∣∣∣∣∣∣ =1 · (−1)1+2

∣∣∣∣∣∣1 −1 1−1 2 0−1 5 3

∣∣∣∣∣∣+ 0 · (−1)2+2

∣∣∣∣∣∣−1 2 2−1 2 0−1 5 3

∣∣∣∣∣∣+3 · (−1)3+2

∣∣∣∣∣∣−1 2 21 −1 1−1 5 3

∣∣∣∣∣∣+ 0 · (−1)4+2

∣∣∣∣∣∣−1 2 21 −1 1−1 2 0

∣∣∣∣∣∣= (−1)[6 + 0− 5− (−2 + 0 + 3)]− 3[3− 2 + 10− (2− 5 + 6)]

= (−1)[1− 1]− 3[11− 3] = −24




∣∣∣∣∣∣∣∣−1 1 2 21 0 −1 1−1 3 2 0−1 0 5 3

∣∣∣∣∣∣∣∣ =1 · (−1)1+2

∣∣∣∣∣∣1 −1 1−1 2 0−1 5 3

∣∣∣∣∣∣+ 0 · (−1)2+2

∣∣∣∣∣∣−1 2 2−1 2 0−1 5 3

∣∣∣∣∣∣+3 · (−1)3+2

∣∣∣∣∣∣−1 2 21 −1 1−1 5 3

∣∣∣∣∣∣+ 0 · (−1)4+2

∣∣∣∣∣∣−1 2 21 −1 1−1 2 0

∣∣∣∣∣∣= (−1)[6 + 0− 5− (−2 + 0 + 3)]− 3[3− 2 + 10− (2− 5 + 6)]

= (−1)[1− 1]− 3[11− 3] = −24



Vypocet determinantu matic radu n ≥ 4

Vypocet determinantu matic n-teho radu, kde n ≥ 4, prevadıme pomocıLaplaceova rozvoje na vypocet determinantu radu (n− 1).

Predtım je vsak vyhodne puvodnı determinant upravit pomocı vlastnostıdeterminantu c. 1 a c. 3 tak, aby v nekterem radku nebo sloupci byly vsechnyprvky krome jednoho rovny nule.

Podle tohoto radku nebo sloupce (s jednım nenulovym prvkem) pakdeterminant rozvineme.

Obdrzıme tak pouze jediny determinant radu (n− 1), ktery opet obdobnymzpusobem upravıme a rozvineme.

Takto postupujeme, az dostaneme determinant 3. radu, ktery jiz vycıslımeSarrusovym pravidlem.






























Prıklad (Vypocet determinantu matice radu 4)∣∣∣∣∣∣∣∣2 1 −1 23 3 −2 1

−2 -1 3 2−3 1 2 −1

∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−|·3

+

=

∣∣∣∣∣∣∣∣0 0 2 4−3 0 7 7

−2 -1 3 2−5 0 5 1

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)2+3

∣∣∣∣∣∣0 2 4−3 7 7−5 5 1

∣∣∣∣∣∣ = (+1)[0− 70− 60− (−140 + 0− 6)] = 16



−2 -1 3 2−3 1 2 −1

∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−|·3

+

=

∣∣∣∣∣∣∣∣0 0 2 4−3 0 7 7

−2 -1 3 2−5 0 5 1

∣∣∣∣∣∣∣∣

=

= (−1) · (−1)2+3

∣∣∣∣∣∣0 2 4−3 7 7−5 5 1

∣∣∣∣∣∣ = (+1)[0− 70− 60− (−140 + 0− 6)] = 16



−2 -1 3 2−3 1 2 −1

∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−|·3

+

=

∣∣∣∣∣∣∣∣0 0 2 4−3 0 7 7

−2 -1 3 2−5 0 5 1

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)2+3

∣∣∣∣∣∣0 2 4−3 7 7−5 5 1

∣∣∣∣∣∣

= (+1)[0− 70− 60− (−140 + 0− 6)] = 16



−2 -1 3 2−3 1 2 −1

∣∣∣∣∣∣∣∣←−+

←−−−−+

←−−−|·3

+

=

∣∣∣∣∣∣∣∣0 0 2 4−3 0 7 7

−2 -1 3 2−5 0 5 1

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)2+3

∣∣∣∣∣∣0 2 4−3 7 7−5 5 1

∣∣∣∣∣∣ = (+1)[0− 70− 60− (−140 + 0− 6)] = 16


Prıklad (Vypocet determinantu matic radu n ≥ 4)∣∣∣∣∣∣∣∣∣∣1 1 1 1 01 1 1 0 11 1 0 1 11 0 1 1 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣ = (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4



∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣

=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣ = (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4



∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣ = (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4



∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣

=

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣ = (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4



∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣

= (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4



∣∣∣∣∣∣∣∣∣∣←−−1+

←−−−−−

−1

+

←−−−−−−−−−

−1

+

=

∣∣∣∣∣∣∣∣∣∣1 1 1 1 00 0 0 −1 10 0 −1 0 10 −1 0 0 10 1 1 1 1

∣∣∣∣∣∣∣∣∣∣=

= 1 · (−1)1+1

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1−1 0 0 11 1 1 1

∣∣∣∣∣∣∣∣ ←−+

=

∣∣∣∣∣∣∣∣0 0 −1 10 −1 0 1

-1 0 0 10 1 1 2

∣∣∣∣∣∣∣∣ =

= (−1) · (−1)3+1

∣∣∣∣∣∣0 −1 1−1 0 11 1 2

∣∣∣∣∣∣ = (−1)[0−1−1− (0+0+2)] = (−1)(−4) = 4


Inverznı matice

Definice (regularnı a singularnı matice)

Ctvercova matice A radu n se nazyva regularnı, jestlize determinant teto maticeje ruzny od nuly. V opacnem prıpade mluvıme o singularnı matici.

Poznamka (regularnı a singularnı matice)

Je-li matice A regularnı, potom

h(A) = n,

radky (sloupce) matice A jsou linearne nezavisle.

Je-li matice A singularnı, potom

h(A) < n,

radky (sloupce) matice A jsou linearne zavisle.


Inverznı matice





h(A) = n,



h(A) < n,



Inverznı matice





h(A) = n,



h(A) < n,



Inverznı matice

Definice (inverznı matice)

Necht’ A je ctvercova matice radu n. Jestlize existuje ctvercova matice A−1

stejneho radu takova, ze platı

A ·A−1 = A−1 ·A = I,

nazyvame matici A−1 inverznı maticı k matici A.

Veta (inverznı matice)

K regularnı matici A existuje prave jedna inverznı matice A−1.

Inverznı matici lze nalezt pouze k regularnı matici, tedy ke ctvercove matici,jejız radky jsou linearne nezavisle.

Inverznı matice je urcena jednoznacne.


Inverznı matice




A ·A−1 = A−1 ·A = I,







Inverznı matice




A ·A−1 = A−1 ·A = I,







Inverznı matice




A ·A−1 = A−1 ·A = I,







Poznamka (vypocet inverznı matice)

1 Transformace rozsırene matice (A|I) na tvar (I|A−1) pomocı ekvivalentnıchradkovych uprav.

2 Pomocı adjungovane matice.

Postup vypoctu prevodem na jednotkovou matici1 Matici A rozsırıme vpravo o jednotkovou matici stejneho radu: (A|I).2 Pomocı ekvivalentnıch radkovych uprav prevedeme matici A na diagonalnı a

pak na jednotkovou matici I. Vsechny upravy provadıme s celymi radkymatice (A|I).

3 Jestlize jsme matici A prevedli na jednotkovou matici I, pak matici I jsmeprevedli na A−1 :

(A|I) ∼ · · · ∼ (I|A−1)4 Pokud matici A na jednotkovou matici nelze prevest (behem vypoctu

dostaneme v leve casti cely radek nulovy), inverznı matice A−1 neexistuje amatice A je singularnı.














Postup vypoctu prevodem na jednotkovou matici1 Matici A rozsırıme vpravo o jednotkovou matici stejneho radu: (A|I).

2 Pomocı ekvivalentnıch radkovych uprav prevedeme matici A na diagonalnı apak na jednotkovou matici I. Vsechny upravy provadıme s celymi radkymatice (A|I).




















(A|I) ∼ · · · ∼ (I|A−1)

4 Pokud matici A na jednotkovou matici nelze prevest (behem vypoctudostaneme v leve casti cely radek nulovy), inverznı matice A−1 neexistuje amatice A je singularnı.











Prıklad (vypocet inverznı matice pr. 1)

Najdete inverznı matici k matici AAA =

(2 71 4

)(

2 7

1 4

∣∣∣∣ 1 00 1

)←−←− ∼

(1 42 7

∣∣∣∣ 0 11 0

)←−−2+∼(1 4

0 -1

∣∣∣∣ 0 11 −2

)←−

4

+

∼(1 00 −1

∣∣∣∣ 4 −71 −2

)| · (−1) ∼

(1 00 1

∣∣∣∣ 4 −7−1 2

)

AAA−1 =

(4 −7−1 2

)



K matici A =

1 −1 2−1 0 1−2 1 2

urcete inverznı matici.

1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/31 0 0

0 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici A =

1 −1 2−1 0 1−2 1 2


1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/31 0 0

0 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici A =

1 −1 2−1 0 1−2 1 2


1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/31 0 0

0 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici A =

1 −1 2−1 0 1−2 1 2


1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/3

1 0 00 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici A =

1 −1 2−1 0 1−2 1 2


1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/31 0 0

0 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici A =

1 −1 2−1 0 1−2 1 2


1 −1 2−1 0 1−2 1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−+

←−−−−

2

+

∼

1 −1 2

0 -1 30 −1 6

∣∣∣∣∣∣1 0 01 1 02 0 1

←−−1+

←−−−−−−1+

∼

1 0 −10 −1 3

0 0 3

∣∣∣∣∣∣0 −1 01 1 01 −1 1

←−−1

+

| · 3 ←−+

∼

3 0 00 −1 00 0 3

∣∣∣∣∣∣1 −4 10 2 −11 −1 1

| · 1/3| · (−1)| · 1/31 0 0

0 1 00 0 1

∣∣∣∣∣∣1/3 −4/3 1/30 −2 11/3 −1/3 1/3

⇒ A−1 =1

3

1 −4 10 −6 31 −1 1



K matici B =

1 2 32 −1 13 −1 2


1 2 32 −1 13 −1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−−2+

←−−−−−

−3

+

∼

1 2 3

0 -5 −50 −7 −7

∣∣∣∣∣∣1 0 0−2 1 0−3 0 1

| 5 ←−

−7+

∼

1 2 30 −5 −50 0 0

∣∣∣∣∣∣1 0 0−2 1 0−1 −7 5

⇒ h(B) = 2

Matice B je singularnı, proto inverznı matice B−1 neexistuje.



K matici B =

1 2 32 −1 13 −1 2


1 2 32 −1 13 −1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−−2+

←−−−−−

−3

+

∼

1 2 3

0 -5 −50 −7 −7

∣∣∣∣∣∣1 0 0−2 1 0−3 0 1

| 5 ←−

−7+

∼

1 2 30 −5 −50 0 0

∣∣∣∣∣∣1 0 0−2 1 0−1 −7 5

⇒ h(B) = 2




K matici B =

1 2 32 −1 13 −1 2


1 2 32 −1 13 −1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−−2+

←−−−−−

−3

+

∼

1 2 3

0 -5 −50 −7 −7

∣∣∣∣∣∣1 0 0−2 1 0−3 0 1

| 5 ←−

−7+

∼

1 2 30 −5 −50 0 0

∣∣∣∣∣∣1 0 0−2 1 0−1 −7 5

⇒ h(B) = 2




K matici B =

1 2 32 −1 13 −1 2


1 2 32 −1 13 −1 2

∣∣∣∣∣∣1 0 00 1 00 0 1

←−−2+

←−−−−−

−3

+

∼

1 2 3

0 -5 −50 −7 −7

∣∣∣∣∣∣1 0 0−2 1 0−3 0 1

| 5 ←−

−7+

∼

1 2 30 −5 −50 0 0

∣∣∣∣∣∣1 0 0−2 1 0−1 −7 5

⇒ h(B) = 2



Adjungovana matice

Definice (adjungovana matice)

Necht’ A je ctvercova matice radu n. Maticı adjungovanou k matici A nazvemematici, jejız prvky se rovnajı algebraickym doplnkum prvku matice AT . Maticiadjungovanou znacıme adjA.

Algebraicke doplnky jsou k prvkum transponovane matice AT .

Poznamka (adjungovana matice)

Adjungovana matice ma tvar

adjA =

AT

11 AT12 · · · AT

1n

AT21 AT

22 · · · AT2n

......

. . ....

ATn1 AT

n2 · · · ATnn

kde AT

ij je algebraicky doplnek k prvku aij matice AT .


Adjungovana matice






adjA =

AT

11 AT12 · · · AT

1n

AT21 AT

22 · · · AT2n

......

. . ....

ATn1 AT

n2 · · · ATnn

kde AT



Adjungovana matice






adjA =

AT

11 AT12 · · · AT

1n

AT21 AT

22 · · · AT2n

......

. . ....

ATn1 AT

n2 · · · ATnn

kde AT



Prıklad (adjungovana matice)

Urcete adjungovanou matici k matici A =

2 1 −1−1 0 23 −1 −2

.

AT =

2 −1 31 0 −1−1 2 −2

AT

11 = (−1)2∣∣∣∣ 0 −1

2 −2

∣∣∣∣ = 2 AT21 = (−1)3

∣∣∣∣ −1 32 −2

∣∣∣∣ = 4

AT12 = (−1)3

∣∣∣∣ 1 −1−1 −2

∣∣∣∣ = 3 AT22 = (−1)4

∣∣∣∣ 2 3−1 −2

∣∣∣∣ = −1AT

13 = (−1)4∣∣∣∣ 1 0−1 2

∣∣∣∣ = 2 AT23 = (−1)5

∣∣∣∣ 2 −1−1 2

∣∣∣∣ = −3AT

31 = (−1)4∣∣∣∣ −1 3

0 −1

∣∣∣∣ = 1

AT32 = (−1)5

∣∣∣∣ 2 31 −1

∣∣∣∣ = 5

AT33 = (−1)6

∣∣∣∣ 2 −11 0

∣∣∣∣ = 1

adjA =

2 3 24 −1 −31 5 1




2 1 −1−1 0 23 −1 −2

.

AT =

2 −1 31 0 −1−1 2 −2

AT11 = (−1)2

∣∣∣∣ 0 −12 −2

∣∣∣∣ = 2 AT21 = (−1)3

∣∣∣∣ −1 32 −2

∣∣∣∣ = 4

AT12 = (−1)3

∣∣∣∣ 1 −1−1 −2

∣∣∣∣ = 3 AT22 = (−1)4

∣∣∣∣ 2 3−1 −2

∣∣∣∣ = −1AT

13 = (−1)4∣∣∣∣ 1 0−1 2

∣∣∣∣ = 2 AT23 = (−1)5

∣∣∣∣ 2 −1−1 2

∣∣∣∣ = −3AT

31 = (−1)4∣∣∣∣ −1 3

0 −1

∣∣∣∣ = 1

AT32 = (−1)5

∣∣∣∣ 2 31 −1

∣∣∣∣ = 5

AT33 = (−1)6

∣∣∣∣ 2 −11 0

∣∣∣∣ = 1

adjA =

2 3 24 −1 −31 5 1




2 1 −1−1 0 23 −1 −2

.

AT =

2 −1 31 0 −1−1 2 −2

AT

11 = (−1)2∣∣∣∣ 0 −1

2 −2

∣∣∣∣ = 2 AT21 = (−1)3

∣∣∣∣ −1 32 −2

∣∣∣∣ = 4

AT12 = (−1)3

∣∣∣∣ 1 −1−1 −2

∣∣∣∣ = 3 AT22 = (−1)4

∣∣∣∣ 2 3−1 −2

∣∣∣∣ = −1AT

13 = (−1)4∣∣∣∣ 1 0−1 2

∣∣∣∣ = 2 AT23 = (−1)5

∣∣∣∣ 2 −1−1 2

∣∣∣∣ = −3

AT31 = (−1)4

∣∣∣∣ −1 30 −1

∣∣∣∣ = 1

AT32 = (−1)5

∣∣∣∣ 2 31 −1

∣∣∣∣ = 5

AT33 = (−1)6

∣∣∣∣ 2 −11 0

∣∣∣∣ = 1

adjA =

2 3 24 −1 −31 5 1




2 1 −1−1 0 23 −1 −2

.

AT =

2 −1 31 0 −1−1 2 −2

AT

11 = (−1)2∣∣∣∣ 0 −1

2 −2

∣∣∣∣ = 2 AT21 = (−1)3

∣∣∣∣ −1 32 −2

∣∣∣∣ = 4

AT12 = (−1)3

∣∣∣∣ 1 −1−1 −2

∣∣∣∣ = 3 AT22 = (−1)4

∣∣∣∣ 2 3−1 −2

∣∣∣∣ = −1AT

13 = (−1)4∣∣∣∣ 1 0−1 2

∣∣∣∣ = 2 AT23 = (−1)5

∣∣∣∣ 2 −1−1 2

∣∣∣∣ = −3AT

31 = (−1)4∣∣∣∣ −1 3

0 −1

∣∣∣∣ = 1

AT32 = (−1)5

∣∣∣∣ 2 31 −1

∣∣∣∣ = 5

AT33 = (−1)6

∣∣∣∣ 2 −11 0

∣∣∣∣ = 1

adjA =

2 3 24 −1 −31 5 1




2 1 −1−1 0 23 −1 −2

.

AT =

2 −1 31 0 −1−1 2 −2

AT

11 = (−1)2∣∣∣∣ 0 −1

2 −2

∣∣∣∣ = 2 AT21 = (−1)3

∣∣∣∣ −1 32 −2

∣∣∣∣ = 4

AT12 = (−1)3

∣∣∣∣ 1 −1−1 −2

∣∣∣∣ = 3 AT22 = (−1)4

∣∣∣∣ 2 3−1 −2

∣∣∣∣ = −1AT

13 = (−1)4∣∣∣∣ 1 0−1 2

∣∣∣∣ = 2 AT23 = (−1)5

∣∣∣∣ 2 −1−1 2

∣∣∣∣ = −3AT

31 = (−1)4∣∣∣∣ −1 3

0 −1

∣∣∣∣ = 1

AT32 = (−1)5

∣∣∣∣ 2 31 −1

∣∣∣∣ = 5

AT33 = (−1)6

∣∣∣∣ 2 −11 0

∣∣∣∣ = 1

adjA =

2 3 24 −1 −31 5 1


Poznamka (Vypocet inverznı matice pomocı adjungovane matice)

Inverznı matici k regularnı matici muzeme urcit pomocı adjungovane matice nazaklade vztahu

A−1 =1

detA· adjA

Pro matice radu n > 3 je tento postup zdlouhavy, protoze je potreba spocıtatn determinantu radu (n− 1). Proto je vyhodnejsı pocıtat inverznı matici k temtomaticım transformacı matice jednotkove.


Poznamka (Vypocet inverznı matice pomocı adjungovane matice)

Inverznı matici k regularnı matici muzeme urcit pomocı adjungovane matice nazaklade vztahu

A−1 =1

detA· adjA

Pro matice radu n > 3 je tento postup zdlouhavy, protoze je potreba spocıtatn determinantu radu (n− 1). Proto je vyhodnejsı pocıtat inverznı matici k temtomaticım transformacı matice jednotkove.


Prıklad (Vypocet inverznı matice pomocı adjungovane matice pr. 1)

K matici AAA =

(−2 53 −6

)urcete inverznı matici.

det AAA =

∣∣∣∣−2 53 −6

∣∣∣∣ = 12− 15 = −3 6= 0, AAAT =

(−2 35 −6

)AAAT

11 = (−1)1+1 · (−6) = −6, AAAT12 = (−1)1+2 · 5 = −5

AAAT21 = (−1)2+1 · 3 = −3, AAAT

22 = (−1)2+2 · (−2) = −2

Adjungovana a inverznı matice k matici AAA :

adjAAA =

(−6 −5−3 −2

)

AAA−1 = −1

3

(−6 −5−3 −2

)=

25

3

12

3



Urcete inverznı matici k matici A =

0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0 AT =

0 −1 22 2 −11 −1 1

AT

11 = (−1)2∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1AT

31 = (−1)4∣∣∣∣ −1 2

2 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2




0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0

AT =

0 −1 22 2 −11 −1 1

AT

11 = (−1)2∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1AT

31 = (−1)4∣∣∣∣ −1 2

2 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2




0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0 AT =

0 −1 22 2 −11 −1 1

AT11 = (−1)2

∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1AT

31 = (−1)4∣∣∣∣ −1 2

2 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2




0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0 AT =

0 −1 22 2 −11 −1 1

AT

11 = (−1)2∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1

AT31 = (−1)4

∣∣∣∣ −1 22 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2




0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0 AT =

0 −1 22 2 −11 −1 1

AT

11 = (−1)2∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1AT

31 = (−1)4∣∣∣∣ −1 2

2 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2




0 2 1−1 2 −12 −1 1

.

det A =

∣∣∣∣∣∣0 2 1−1 2 −12 −1 1

∣∣∣∣∣∣ = −5 6= 0 AT =

0 −1 22 2 −11 −1 1

AT

11 = (−1)2∣∣∣∣ 2 −1−1 1

∣∣∣∣ = 1 AT21 = (−1)3

∣∣∣∣ −1 2−1 1

∣∣∣∣ = −1AT

12 = (−1)3∣∣∣∣ 2 −1

1 1

∣∣∣∣ = −3 AT22 = (−1)4

∣∣∣∣ 0 21 1

∣∣∣∣ = −2AT

13 = (−1)4∣∣∣∣ 2 2

1 −1

∣∣∣∣ = −4 AT23 = (−1)5

∣∣∣∣ 0 −11 −1

∣∣∣∣ = −1AT

31 = (−1)4∣∣∣∣ −1 2

2 −1

∣∣∣∣ = −3AT

32 = (−1)5∣∣∣∣ 0 2

2 −1

∣∣∣∣ = 4

AT33 = (−1)6

∣∣∣∣ 0 −12 2

∣∣∣∣ = 2

A−1 = −1

5

1 −3 −4−1 −2 −1−3 4 2

=

=1

5

−1 3 41 2 13 −4 −2


Maticove rovnice

PoznamkaRovnice, ve kterych je neznamou matice, se nazyvajı maticove rovnice.

A+O = A, O +A = A

A · I = A, I ·A = A

nasobenı matic nenı komutativnı (platı i pri vytykanı matic)

mısto delenı matic pouzıvame nasobenı inverznı maticı

A ·A−1 = I, A−1 ·A = I

Poznamka

Necht’ A je regularnı matice. Potom

A ·X = B ⇒ X = A−1 ·B (nasobenı zleva inverznı maticı)

X ·A = B ⇒ X = B ·A−1 (nasobenı zprava inverznı maticı)


Maticove rovnice


A+O = A, O +A = A

A · I = A, I ·A = A



A ·A−1 = I, A−1 ·A = I

Poznamka





Maticove rovnice


A+O = A, O +A = A

A · I = A, I ·A = A



A ·A−1 = I, A−1 ·A = I

Poznamka





Prıklad (maticove rovnice pr. 1)

Reste maticovou rovnici X · (AT +B) = 2(X + I) ·B − C, je-li

A =

(0 −23 1

), B =

(2 −3−2 0

), C =

(2 8−2 1

).

X ·AT +X ·B = 2X ·B + 2I ·B − C

X ·AT −X ·B = 2B − C

X · (AT −B) = 2B − C

X · (AT −B) · (AT −B)−1 = (2B − C) · (AT −B)−1

X = (2B − C) · (AT −B)−1

(AT −B) =

(−2 60 1

), (AT −B)−1 =

1

−2

(1 −60 −2

)2B − C =

(2 −14−2 −1

), X =

1

2

(−2 −162 −14

)=

(−1 −81 −7

)




A =

(0 −23 1

), B =

(2 −3−2 0

), C =

(2 8−2 1

).

X ·AT +X ·B = 2X ·B + 2I ·B − C

X ·AT −X ·B = 2B − C

X · (AT −B) = 2B − C

X · (AT −B) · (AT −B)−1 = (2B − C) · (AT −B)−1

X = (2B − C) · (AT −B)−1

(AT −B) =

(−2 60 1

), (AT −B)−1 =

1

−2

(1 −60 −2

)2B − C =

(2 −14−2 −1

), X =

1

2

(−2 −162 −14

)=

(−1 −81 −7

)




A =

(0 −23 1

), B =

(2 −3−2 0

), C =

(2 8−2 1

).

X ·AT +X ·B = 2X ·B + 2I ·B − C

X ·AT −X ·B = 2B − C

X · (AT −B) = 2B − C

X · (AT −B) · (AT −B)−1 = (2B − C) · (AT −B)−1

X = (2B − C) · (AT −B)−1

(AT −B) =

(−2 60 1

), (AT −B)−1 =

1

−2

(1 −60 −2

)

2B − C =

(2 −14−2 −1

), X =

1

2

(−2 −162 −14

)=

(−1 −81 −7

)




A =

(0 −23 1

), B =

(2 −3−2 0

), C =

(2 8−2 1

).

X ·AT +X ·B = 2X ·B + 2I ·B − C

X ·AT −X ·B = 2B − C

X · (AT −B) = 2B − C

X · (AT −B) · (AT −B)−1 = (2B − C) · (AT −B)−1

X = (2B − C) · (AT −B)−1

(AT −B) =

(−2 60 1

), (AT −B)−1 =

1

−2

(1 −60 −2

)2B − C =

(2 −14−2 −1

), X =

1

2

(−2 −162 −14

)=

(−1 −81 −7

)



Reste maticovou rovnici 2A+A ·X −X = AT + 2X, je-li

A =

(4 −25 3

).

2A+A ·X −X = AT + 2X

A ·X − 3X = AT − 2A

(A− 3I) ·X = AT − 2A

(A− 3I)−1 · (A− 3I) ·X = (A− 3I)−1 · (AT − 2A)

X = (A− 3I)−1 · (AT − 2A)

A− 3I =

(1 −25 0

), (A− 3I)−1 =

1

10

(0 2−5 1

)AT − 2A =

(−4 9−12 −3

), X =

1

10

(−24 −6

8 −48

)=

1

5

(−12 −3

4 −24

)




A =

(4 −25 3

).

2A+A ·X −X = AT + 2X

A ·X − 3X = AT − 2A

(A− 3I) ·X = AT − 2A

(A− 3I)−1 · (A− 3I) ·X = (A− 3I)−1 · (AT − 2A)

X = (A− 3I)−1 · (AT − 2A)

A− 3I =

(1 −25 0

), (A− 3I)−1 =

1

10

(0 2−5 1

)AT − 2A =

(−4 9−12 −3

), X =

1

10

(−24 −6

8 −48

)=

1

5

(−12 −3

4 −24

)




A =

(4 −25 3

).

2A+A ·X −X = AT + 2X

A ·X − 3X = AT − 2A

(A− 3I) ·X = AT − 2A

(A− 3I)−1 · (A− 3I) ·X = (A− 3I)−1 · (AT − 2A)

X = (A− 3I)−1 · (AT − 2A)

A− 3I =

(1 −25 0

), (A− 3I)−1 =

1

10

(0 2−5 1

)

AT − 2A =

(−4 9−12 −3

), X =

1

10

(−24 −6

8 −48

)=

1

5

(−12 −3

4 −24

)




A =

(4 −25 3

).

2A+A ·X −X = AT + 2X

A ·X − 3X = AT − 2A

(A− 3I) ·X = AT − 2A

(A− 3I)−1 · (A− 3I) ·X = (A− 3I)−1 · (AT − 2A)

X = (A− 3I)−1 · (AT − 2A)

A− 3I =

(1 −25 0

), (A− 3I)−1 =

1

10

(0 2−5 1

)AT − 2A =

(−4 9−12 −3

), X =

1

10

(−24 −6

8 −48

)=

1

5

(−12 −3

4 −24

)



Reste maticovou rovnici AAAXXXBBB = CCC, kde

AAA =

(−3 5−2 4

), BBB =

(4 35 4

), CCC =

(−2 4−2 6

)

AAAXXXBBB = CCC

AAA−1AAAXXXBBB = AAA−1CCC

XXXBBB = AAA−1CCC

XXXBBBBBB−1 = AAA−1CCCBBB−1

XXX = AAA−1CCCBBB−1

AAA−1 =1

2

(−4 5−2 3

), BBB−1 =

(4 −3−5 4

)XXX =

1

2

(−4 5−2 3

)(−2 4−2 6

)(4 −3−5 4

)=

1

2

(−2 14−2 10

)(4 −3−5 4

)=

=

(−39 31−29 23

)c©Dana Rıhova (Mendelu Brno) Determinanty 38 / 38

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