linearna_algebra -- Sage

attach DATA+'LinearnaAlgebra.sage'

U paketu se nalaze trenutno tri funkcije:
bazicna_rjesenja - daje bazicna rjesenja sustava linearnih jednadzbi
linear_solve - rjesava sustav zapisan u matricnom obliku. Moguce je
birati parametre po zelji.
solve_mat - rjesava matricne jednadzbe tako da se ne moraju
raspisivati u sustave linearnih jednadzbi

x,y,z,u=var('x y z u')

bazicna_rjesenja([[1,1,2,3],[1,-4,2,-3],[0,7,2,5],[1,3,4,2]],[6,-4,12,8],[x,y,z,u])

[x == 0, y == (8/13), z == (25/26), u == (15/13)]
[y == 0, x == (-8/3), z == (11/6), u == (5/3)]
[z == 0, x == (50/17), y == (22/17), u == (10/17)]
[u == 0, x == 6, y == 2, z == -1]

bazicna_rjesenja(matrix([[1,1,2,3],[1,-4,2,-3],[0,7,2,5],[1,3,4,2]]),[6,-4,12,8],[x,y,z,u])

[x == 0, y == (8/13), z == (25/26), u == (15/13)]
[y == 0, x == (-8/3), z == (11/6), u == (5/3)]
[z == 0, x == (50/17), y == (22/17), u == (10/17)]
[u == 0, x == 6, y == 2, z == -1]

bazicna_rjesenja([[1,1,2,3],[1,-4,2,-3],[0,7,2,5],[1,3,4,2]],matrix(4,1,[6,-4,12,8]),[x,y,z,u])

[x == 0, y == (8/13), z == (25/26), u == (15/13)]
[y == 0, x == (-8/3), z == (11/6), u == (5/3)]
[z == 0, x == (50/17), y == (22/17), u == (10/17)]
[u == 0, x == 6, y == 2, z == -1]

bazicna_rjesenja([[2,3],[-1,5]],[1,-7],[x,y])

 $\newcommand{\Bold}[1]{\mathbf{#1}}\verb|Error:|\phantom{\verb!x!}\verb|sustav|\phantom{\verb!x!}\verb|ima|\phantom{\verb!x!}\verb|jedinstveno|\phantom{\verb!x!}\verb|rjesenje|$

bazicna_rjesenja([[1,-4,5],[0,-3,2],[2,7,0]],[6,-12,35],[x,y,z])

 $\newcommand{\Bold}[1]{\mathbf{#1}}\verb|Error:|\phantom{\verb!x!}\verb|sustav|\phantom{\verb!x!}\verb|je|\phantom{\verb!x!}\verb|kontradiktoran|$

x1,x2,x3,x4=var('x1 x2 x3 x4')

A=matrix(3,4,[2,3,2,6,-2,3,-6,12,2,6,0,15]) B=matrix(3,1,[1,-19,-8])

linear_solve(A,B,[x1,x2,x3,x4],[x1,x2,x4,x3])

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[x_{1} = -2 \, r_{1} + \frac{3}{2} \, r_{2} + 5, x_{2} = \frac{2}{3} \, r_{1} - 3 \, r_{2} - 3, x_{4} = r_{2}, x_{3} = r_{1}\right]\right]$

linear_solve(A,B,[x1,x2,x3,x4],[x2,x4,x3,x1])

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[x_{2} = -2 \, r_{3} - \frac{10}{3} \, r_{4} + 7, x_{4} = \frac{2}{3} \, r_{3} + \frac{4}{3} \, r_{4} - \frac{10}{3}, x_{3} = r_{4}, x_{1} = r_{3}\right]\right]$

linear_solve(A,B,[x1,x2,x3,x4],[x2,x4,x3,x1],rjecnik=True)

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left\{x_{1} : r_{5}, x_{4} : \frac{2}{3} \, r_{5} + \frac{4}{3} \, r_{6} - \frac{10}{3}, x_{3} : r_{6}, x_{2} : -2 \, r_{5} - \frac{10}{3} \, r_{6} + 7\right\}\right]$

var('a b c d') A=matrix([[1,1],[0,1]]) X=matrix([[a,b],[c,d]])

solve_mat(A*X,X*A,(a,b,c,d))

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[a = r_{7}, b = r_{8}, c = 0, d = r_{7}\right]\right]$

solve_mat(A*X,X*A,(a,b,c,d),rjecnik=True)

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left\{d : r_{9}, c : 0, b : r_{10}, a : r_{9}\right\}\right]$

X.subs(d=a,c=0)

 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} a & b \\ 0 & a \end{array}\right)$