1. OdreditirangmatricaA,B,A+BiABakojeA =__2 1 12 2 11 0 0__ i B
=__2 2 22 0 11 1 1__Resenje:RangmatriceA:A =__2 1 12 2 11 0 0__ __2
1 10 3 01 0 0__ __1 1 20 3 00 0 1__Proizvodnaglavnojdijagonalije1 3
(1) = 3 = 0 (A) = 3.RangmatriceB:B =__2 2 22 0 11 1 1__ __1 1 12 0
12 2 2__ __1 1 12 0 10 4 0__ __1 1 10 2 30 4 0__ __1 1 10 3 20 0
4__Proizvodnaglavnojdijagonalije 1 3 (4) = 13 = 0 (B) =
3.RangmatriceA+B:A+B =__2 1 12 2 11 0 0__ +__2 2 22 0 11 1 1__ =__4
3 34 2 22 1 1__ __4 3 34 2 22 1 1__ __4 3 32 1 14 2 2__ __4 3 32 1
10 0 0__det(A+B) = 0ali4 32 1 = 4 6 = 10 = 0 (A+B) = 2RangmatriceA
B:A B =__2 1 12 2 11 0 0__ __2 2 22 0 11 1 1__ =__1 5 27 5 52 2 2__
__1 5 27 5 52 2 2__ __1 5 20 30 90 8 6__ __1 5 20 30 90 0
2.8__Proizvodnaglavnojdijagonalije 3 2 (2) = 12 = 0 (A B) = 312.
OdreditirangmatriceA =__a a 2 02a a a 1 13a 3a 3 a 1
a__,uzavisnostiodvrednostirealnogparametraa.Resenje:A =__a a 2 02a
a a 1 13a 3a 3 a 1 a__Zaa = 0,A =__0 0 2 00 0 1 10 0 3 1__ (A) =
2Zaa = 0 (A) = 33. Resitimatricnujednacinu(AX)1+X1= BakojeA =__2 0
00 1 00 0 3__ i B =__1 3 21 1 11 2
1__.Resenje:Najpreresavamozadatumatricnujednacinu:(AX)1+X1= B(X)1
(A)1+X1= BX1 (A1+I) = B / XsalevestraneX X1..I(A1+I) = X B(A1+I) =
X B / B1sadesnestrane(A1+I) B1= X B B1. .IX = (A1+I)
B1TrazimoinverznumatricumatriceA:A1=adjA|A|detA =2 0 00 1 00 0 3= 2
(1 (3) 0) =
6,matricajeregularnaTrazimokofaktoreiadugovanumatricu:adjA
=__A11A21A31A12A22A32A13A23A33__2A11=(1)1+11 00 3=1 (3) = 3;
A21=(1)2+10 00 3=1 0 = 0;A31=(1)3+10 01 0=1 (1) = 0;A12=(1)1+20 00
3=1 0 = 0; A22=(1)2+22 00 3=1 (6) = 6;A32=(1)3+22 00 0=1 0 =
0;A13=(1)1+30 10 0=1 0 = 0; A23=(1)2+32 00 0=1 0 = 0;A33=(1)3+32 00
1=1 2 = 2;adjA=__3 0 00 6 00 0 2__A1=16 __3 0 00 6 00 0 2__ =__1/2
0 00 1 00 0 1/3__B1=adjB|B|TrazimoinverznumatricumatriceB:detB =1 3
21 1 11 2 1= 1 (1 1 + 2 1) + 3 (1 1 1 1) + 2 (2 1) =
3,matricajeregularna.Trazimokofaktoreiadugovanumatricu:adjB
=__B11B21B31B12B22B32B13B23B33__B11= (1)1+11 12 1 = 1 (1 2) =
1;B21= (1)2+13 22 1 = 1 (3 + 4) = 1;B31= (1)3+13 21 1 = 1 (3 2) =
1;B12= (1)1+21 11 1 = 1 (1 + 1) = 2;B22= (1)2+21 21 1 = 1 (1 2) =
1;B32= (1)3+21 21 1 = 1 (1 2) = 3;B13= (1)1+31 11 2 = 1 (2 1) =
3;3B23=(1)2+31 31 2 = 1 (2 + 3) = 1;B33= (1)3+31 21 1 = 1 (1 + 3) =
4;adjB =__1 1 12 1 33 1 4__B1= 1 __1 1 12 1 33 1 4__ =__1 1 12 1 33
1 4__X = (A1+I) B1_____1/2 0 00 1 00 0 1/3__ +__1 0 00 1 00 0
1_______1 1 12 1 33 1 4__ =__3/2 3/2 3/24 2 62 2/3 8/3__4.
KoristeciKramerovopravilo,zaa, c
R,resitisistemlinearnihjednacinax+y+z= 3xy+z= a+2cx+yz=
aResenje:Determinanatasistemaje:D =1 1 11 1 11 1 1= 1 (1 1) 1 (1 1)
+ 1 (1 + 1) = 4Dx=3a 1 1a + 2c 1 1a 1 1= 3a (1 1) (a + 2c) (1 1) +
a (1 + 1) = 4 (a + c)Dy=1 3a 11 a + 2c 11 a 1= 3a (1 1) (a + 2c) (1
1) a (1 1) = 4 (a c)Dz=1 1 a1 1 a + 2c1 1 a= 3a (1 + 1) (a + 2c) (1
1) + a (1 1) =
4aJedinstvenaresenjasistema,odnosnovrednostizanepoznatex, y,
zsusledeca:x =DxD=4 (a + c)4= a + cy=DyD=4 (a c)4= a cz=DzD=4a4=
a(x, y, z) = (a + c, a c, a)45.
Diskutovatiresivostsistemalineranihjednacina2x+6y+(a + 6)z= 0x+7y+
5z= 0ax+y z=
0uzavisnostiodvrednostirealnogparametraa.Resenje:RadiseohomogenomsistemuRangovematricemorabitimanjiod3,dabisistemimaonetrivijalanaresenja.A
=__2 6 a + 61 7 5a 5 13__ __1 7 52 6 a + 6a 5 13__ __1 7 50 20 a +
160 5 + 7a 13 + 5a__ __1 7 50 20 a + 160 07a217a +
18020__naovommestumorabitinula7a217a + 18020= 07a217a + 180 =
0a1,2=17 1724 (7) 18020 a1= 4 a2=
457Postojedvevrednostizaparametarazakojesistemimanetrivijalnaresenja.Proucicedmoobemogucnosti.Zaa
= 4__1 7 50 20 a + 160 07a217a + 18020____1 7 50 20 200 0
0__Izdrugevrsteimamo20y + 20z= 0 y= zIzprvevrsteimamo x + 7y + 5z=
0 x = 2zResenjeje: (x, y, z) = (2z, z, z)Zaa = 457__1 7 50 20 a +
160 07a217a + 18020____1 7 50 20 67/70 0 0__Izdrugevrsteimamo20y
+677z= 0 y= 67140zIzprvevrsteimamo x + 7y + 5z= 0 x
=3320zResenjeje: (x, y, z) = (3320z, 67140z, z)5Zakljucak:Zaa =
457; a = 4sistemimasamotrivijalanaresenja(0, 0, 0)Zaa =
457sistemimaresenja(x, y, z) = (3320z, 67140z, z)Zaa =
4sistemimaresenja(x, y, z) = (2z, z, z)6.
Resitisistemlinearnihjednacina2x+yz= 24xy3z = 03x+3y+3z=
2koristecimatricnizapis.Resenje:Matricaprosirenogsistemaje:__2 1 1
24 1 3 03 3 2 1__ __2 1 1 20 3 1 6092322____2 1 1 20 3 1 60 0 0
7__Redukovanisistemjednacinaje:2x+yz= 23yz = 00 =
7Poslednjajednacinasadrziprotivrecnostpajesistemprotivrecan.7.
KoristeciKroneker-Kapelijevstav,ispitatiresivostsistemalinearnihjednacina2x1+x2x3=
2x1+x2+2x3= 33x1+3x2x3=
1Resenje:KronerKapelijevateorema:Sistemimaresenjaakoi samoakoje
rangmatrice sistemajednakmatrici prosirenogsistema,tj.(A) =
(A)Matricasistemaje:A =__2 1 11 1 21 2
1__amatricaprosirenogsistemaje:A =__2 1 1 21 1 2 01 2 1
1__6Umatriciprosirenogsistemapravimonule:A =__2 1 1 21 1 2 31 2 1
1__ __1 2 1 11 1 2 32 1 1 2__ __1 2 1 10 3 1 40 3 1 4__ __1 2 1 10
3 1 40 0 2 0__2x3= 0, x3= 03x2 + x3= 4, x3=43x1 + 2x2x3= 1, x1= 1 2
43= 538. Odreditisopstvenuvrednostmatrice__2 8 121 4 40 0
1__,zatimzapozitivnesopstvenevrednostiodreditisopstvenevektore.Resenje:Zazadatumatricukarakteristicanpolinomsedobijaresavanjemdeterminante:|AE|
=2 8 121 4 40 0 1 = (1)((2)(4)8) = (1)(2216)(1 ) (22 16) = 01 = 0
22 16 = 01= 1 2/3=2 682, 2= 1 +17 3= 1 17Zasopstvenuvrednost =
1dobijasehomogenisistem:3x+8y12z= 0x+3y+4z= 0/30= 03x+8y12z=
03x+9y+12z= 017y= 0 y= 03x = 12z x = 4zKarakteristicnivektorza =
1je:__xyz__ =__4z0z__ = z __401__Zasopstvenuvrednost = 1
17dobijasehomogenisistem:(3 +17)x + 8y 12z= 0x + (3 17)y + 4z=
0717z= 0Iztrecejednacinedobijamoz= 0Izprvejednacinedobijamo (3
+17)x + 8y= 0x = (3 +17)yKarakteristicnivektorza = 1 +17je:__xyz__
=__(3 +17)yy0__ = y __3 +1710__Zasopstvenuvrednost = 1
17dobijasehomogenisistem:(3 +17)x + 8y 12z= 0x + (3 +17)y + 4z=
017z= 0Iztrecejednacinedobijamoz= 0Izprvejednacinedobijamo(3 +17)x
+ 8y= 0x = (3 17)yKarakteristicnivektorza = 1 17je:__xyz__ =__(3
17)yy0__ = y __3 1710__8
