load LRmat1 A A = 3 2 1 2 3 1 2 1 3 5 5 2 b b = 5 1 11 6 % Rozsirena matice soustavy Ab = [A,b] Ab = 3 2 1 5 2 3 1 1 2 1 3 11 5 5 2 6 % % Reseni Gaussovou eliminaci: % =========================== [S,U,np,jb] = GEdemo(Ab) m = 4 n = 4 A = 3 2 1 5 2 3 1 1 2 1 3 11 5 5 2 6 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [4 1] r = 4 jb = 1 k = 1 PREHOZENI k-ho a r-teho RADKU: A = 5 5 2 6 2 3 1 1 2 1 3 11 3 2 1 5 VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 2/5 2/5 3/5 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 5 5 2 6 0 1 1/5 -7/5 0 -1 11/5 43/5 0 -1 -1/5 7/5 Naposledy eliminovany sloupec: s = 1 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [2 2] rs = 2 2 r = 2 jb = 1 2 k = 2 Zamena radku se neprovadi VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: -1 -1 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 5 5 2 6 0 1 1/5 -7/5 0 0 12/5 36/5 0 0 0 0 Naposledy eliminovany sloupec: s = 2 Vlozte polohu dalsiho schodu ve tvaru [r,s]: 3 rs = 3 jb = 1 2 3 S = 5 5 2 6 0 1 1/5 -7/5 0 0 12/5 36/5 0 0 0 0 U = 5 5 2 0 1 1/5 0 0 12/5 np = 1 jb = 1 2 3 % V S neni u posledniho sloupce dodatecny schod, takze soustava je resitelna. % Reseni je jedine, nebot hodnost je 3, coz je prave pocet sloupcu v A, resp. v S % Reseni soustavy s horni trojuhelnikovou matici: % a) pomoci inverze x=inv(U)*S(1:3,4) x = 2 -2 3 % primo zpetnym doszovanim: x3=(36/5)/(12/5) x3 = 3 x2 = -7/5 - (1/5)*x3 x2 = -2 x1 = (6 - 5*x2-2*x3)/5 x1 = 2 % Mame totez jako predtim: [x1;x2;x3] ans = 2 -2 3 diary off % % Totez Jordanovou eliminaci: % =========================== [S,D,np,jb] = JEdemo(Ab) m = 4 n = 4 A = 3 2 1 5 2 3 1 1 2 1 3 11 5 5 2 6 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [4 1] r = 4 jb = 1 k = 1 PREHOZENI k-ho a r-teho RADKU: A = 5 5 2 6 2 3 1 1 2 1 3 11 3 2 1 5 VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 0/0 2/5 2/5 3/5 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 5 5 2 6 0 1 1/5 -7/5 0 -1 11/5 43/5 0 -1 -1/5 7/5 Naposledy eliminovany sloupec: s = 1 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [2 2] rs = 2 2 r = 2 jb = 1 2 k = 2 Zamena radku se neprovadi VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 5 0/0 -1 -1 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 5 0 1 13 0 1 1/5 -7/5 0 0 12/5 36/5 0 0 0 0 Naposledy eliminovany sloupec: s = 2 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [3 3] rs = 3 3 r = 3 jb = 1 2 3 k = 3 Zamena radku se neprovadi VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 5/12 1/12 0/0 0 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 5 0 0 10 0 1 0 -2 0 0 12/5 36/5 0 0 0 0 Naposledy eliminovany sloupec: s = 3 Vlozte polohu dalsiho schodu ve tvaru [r,s]: rs = [] jb = 1 2 3 S = 5 0 0 10 0 1 0 -2 0 0 12/5 36/5 0 0 0 0 D = 5 0 0 0 1 0 0 0 12/5 np = 1 jb = 1 2 3 x1=10/5 x1 = 2 x2=(-2)/1 x2 = -2 x3=(36/5)/(12/5) x3 = 3 diary off load LRmat2 A A = 5 -9 5 2 3 3 1 8 0 1 -2 1 b b = 1 2 1 0 [S,U,np,jb] = GEdemo([A,b]) m = 4 n = 4 A = 5 -9 5 1 2 3 3 2 1 8 0 1 1 -2 1 0 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [4 1] r = 4 jb = 1 k = 1 PREHOZENI k-ho a r-teho RADKU: A = 1 -2 1 0 2 3 3 2 1 8 0 1 5 -9 5 1 VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 2 1 5 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 1 -2 1 0 0 7 1 2 0 10 -1 1 0 1 0 1 Naposledy eliminovany sloupec: s = 1 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [4 2] rs = 4 2 r = 4 jb = 1 2 k = 2 PREHOZENI k-ho a r-teho RADKU: A = 1 -2 1 0 0 1 0 1 0 10 -1 1 0 7 1 2 VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 10 7 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 1 -2 1 0 0 1 0 1 0 0 -1 -9 0 0 1 -5 Naposledy eliminovany sloupec: s = 2 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [3,3] rs = 3 3 r = 3 jb = 1 2 3 k = 3 Zamena radku se neprovadi VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: -1 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 1 -2 1 0 0 1 0 1 0 0 -1 -9 0 0 0 -14 Naposledy eliminovany sloupec: s = 3 Vlozte polohu dalsiho schodu ve tvaru [r,s]: rs = [] jb = 1 2 3 S = 1 -2 1 0 0 1 0 1 0 0 -1 -9 0 0 0 -14 U = 1 -2 1 0 1 0 0 0 -1 np = 2 jb = 1 2 3 % Soustava neni resitelna, nebot S ma ctyri nenulove radky (4=hodnost rozsirene matice [A,b]): S(:,1:3) % ma 3 nenulove radky (3=hodnost A) ans = 1 -2 1 0 1 0 0 0 -1 0 0 0 diary off load LRmat3 A A = 2 -3 2 1 -2 1 5 -9 5 b b = 1 0 1 [S,U,np,jb] = GEdemo([A,b]) m = 3 n = 4 A = 2 -3 2 1 1 -2 1 0 5 -9 5 1 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [2 1] r = 2 jb = 1 k = 1 PREHOZENI k-ho a r-teho RADKU: A = 1 -2 1 0 2 -3 2 1 5 -9 5 1 VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 2 5 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 1 -2 1 0 0 1 0 1 0 1 0 1 Naposledy eliminovany sloupec: s = 1 Vlozte polohu dalsiho schodu ve tvaru [r,s]: [2 2] rs = 2 2 r = 2 jb = 1 2 k = 2 Zamena radku se neprovadi VEKTOR TRANSFORMACNICH SKALARU V s-tem SLOUPCI: 1 ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n: A = 1 -2 1 0 0 1 0 1 0 0 0 0 Naposledy eliminovany sloupec: s = 2 Vlozte polohu dalsiho schodu ve tvaru [r,s]: rs = [] jb = 1 2 S = 1 -2 1 0 0 1 0 1 0 0 0 0 U = 1 -2 0 1 np = 1 jb = 1 2 % y1,y2 jsou vazane, t je volna promenna: % partikularni reseni pro t=0 dostaneme reseni U*y=[0;1]: t = 0 y2 = 1 y2 = 1 y1 = 0 - (-2)*y2 y1 = 2 % Kontrola: U*[y1;y2] ans = 0 1 x0=[y1;y2;t] % Partikularni reseni x0 = 2 1 0 3-2 % dimenze jadra ans = 1 % Baze t = 1 t = 1 % Resime U*y=-[1;0] y2 = 0 y2 = 0 y1 = -1-(-2)*y2 y1 = -1 [y1;y2;t] %Baze ans = -1 0 1 % Vsechna reseni: x0+t*x1 = [2,1,0]+t*[-1,0,1]=[2-t,1,t] diary off