load LUmat1
A
A =
      1            2            1            4      
      2            0            4            3      
      4            2            2            1      
     -3            1            3            2      
[L,U,p,jb] = ludemo(A)
m =
      4      
n =
      4      
A =
      1            2            1            4            1      
      2            0            4            3            2      
      4            2            2            1            3      
     -3            1            3            2            4      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: [3 1]
r =
      3      
jb =
      1      
k =
      1      
PREHOZENI k-ho a r-teho RADKU:
A =
      4            2            2            1            3      
      2            0            4            3            2      
      1            2            1            4            1      
     -3            1            3            2            4      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
      4            2            2            1            3      
     1/2           0            4            3            2      
     1/4           2            1            4            1      
    -3/4           1            3            2            4      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
      4            2            2            1            3      
     1/2          -1            3           5/2           2      
     1/4          3/2          1/2         15/4           1      
    -3/4          5/2          9/2         11/4           4      
Posledni 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 =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/4          3/2          1/2         15/4           1      
     1/2          -1            3           5/2           2      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/4          3/5          1/2         15/4           1      
     1/2         -2/5           3           5/2           2      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/4          3/5        -11/5         21/10          1      
     1/2         -2/5         24/5         18/5           2      
Posledni eliminovany sloupec:
s =
      2      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: [4 3]
rs =
      4            3      
r =
      4      
jb =
      1            2            3      
k =
      3      
PREHOZENI k-ho a r-teho RADKU:
A =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/2         -2/5         24/5         18/5           2      
     1/4          3/5        -11/5         21/10          1      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/2         -2/5         24/5         18/5           2      
     1/4          3/5        -11/24        21/10          1      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
      4            2            2            1            3      
    -3/4          5/2          9/2         11/4           4      
     1/2         -2/5         24/5         18/5           2      
     1/4          3/5        -11/24        15/4           1      
Posledni eliminovany sloupec:
s =
      3      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: 4
rs =
      4      
jb =
      1            2            3            4      
L =
      1            0            0            0      
    -3/4           1            0            0      
     1/2         -2/5           1            0      
     1/4          3/5        -11/24          1      
U =
      4            2            2            1      
      0           5/2          9/2         11/4     
      0            0          24/5         18/5     
      0            0            0          15/4     
p =
      3      
      4      
      2      
      1      
jb =
      1            2            3            4      
A(p,:)  % Permutovana matice
ans =
      4            2            2            1      
     -3            1            3            2      
      2            0            4            3      
      1            2            1            4      
% se rozlozi na L*U:
L*U
ans =
      4            2            2            1      
     -3            1            3            2      
      2            0            4            3      
      1            2            1            4      
load LUmat2
A
A =
      0            1            3           -1            2            1      
      0           -2           -6            2            0            2      
      0            0            0            1           -1            7      
      0            2            6           -2            2            0      
[L,U,p,jb] = ludemo(A)
m =
      4      
n =
      6      
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0           -2           -6            2            0            2      
      0            0            0            1           -1            7      
      0            2            6           -2            2            0      
  Column 7 
      1      
      2      
      3      
      4      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: [1 2]
r =
      1      
jb =
      2      
k =
      1      
PREHOZENI k-ho a r-teho RADKU:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0           -2           -6            2            0            2      
      0            0            0            1           -1            7      
      0            2            6           -2            2            0      
  Column 7 
      1      
      2      
      3      
      4      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0           -2           -6            2            0            2      
      0            0            0            1           -1            7      
      0            2            6           -2            2            0      
  Column 7 
      1      
      2      
      3      
      4      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0           -2            0            0            4            4      
      0            0            0            1           -1            7      
      0            2            0            0           -2           -2      
  Column 7 
      1      
      2      
      3      
      4      
Posledni eliminovany sloupec:
s =
      2      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: [3 4]
rs =
      3            4      
r =
      3      
jb =
      2            4      
k =
      2      
PREHOZENI k-ho a r-teho RADKU:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0           -2           -2      
  Column 7 
      1      
      3      
      2      
      4      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0           -2           -2      
  Column 7 
      1      
      3      
      2      
      4      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0           -2           -2      
  Column 7 
      1      
      3      
      2      
      4      
Posledni eliminovany sloupec:
s =
      4      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: [3 5]
rs =
      3            5      
r =
      3      
jb =
      2            4            5      
k =
      3      
PREHOZENI k-ho a r-teho RADKU:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0           -2           -2      
  Column 7 
      1      
      3      
      2      
      4      
ULOZENI TRANSFORMACNICH SKALARU MISTO NUL DO s-teho SLOUPCE:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0          -1/2          -2      
  Column 7 
      1      
      3      
      2      
      4      
ELIMINACE ZBYVAJICICH SLOUPCU s+1,...,n:
A =
  Columns 1 through 6 
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2            0            0            4            4      
      0            2            0            0          -1/2           0      
  Column 7 
      1      
      3      
      2      
      4      
Posledni eliminovany sloupec:
s =
      5      
Vlozte polohu dalsiho schodu ve tvaru [r,s]: 
rs =
     []
jb =
      2            4            5      
L =
      1            0            0            0      
      0            1            0            0      
     -2            0            1            0      
      2            0          -1/2           1      
U =
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0            0            0            0            4            4      
      0            0            0            0            0            0      
p =
      1      
      3      
      2      
      4      
jb =
      2            4            5      
A(p,:)  % Permutovana matice
ans =
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2           -6            2            0            2      
      0            2            6           -2            2            0      
% se rozlozi na L*U:
L*U
ans =
      0            1            3           -1            2            1      
      0            0            0            1           -1            7      
      0           -2           -6            2            0            2      
      0            2            6           -2            2            0      
diary off