In [2]:

    

import numpy as np
from numpy.linalg import *
rg = matrix_rank
from IPython.display import display, Math, Latex, Markdown


from sympy import *

pr = lambda s: display(Markdown('$'+str(latex(s))+'$'))

def pmatrix(a, intro='',ending='',row=False):
    if len(a.shape) > 2:
        raise ValueError('pmatrix can at most display two dimensions')
    lines = str(a).replace('[', '').replace(']', '').splitlines()
    rv = [r'\begin{pmatrix}']
    rv += ['  ' + ' & '.join(l.split()) + r'\\' for l in lines]
    rv +=  [r'\end{pmatrix}']
    if row:
        return(intro+'\n'.join(rv)+ending)
    else:
        display(Latex('$$'+intro+'\n'.join(rv)+ending+'$$'))


    

In [6]:

    

C = np.array([[1,2],
              [0,1]])
pmatrix(C, intro=r'C_{2\times 2}=')

D = np.array([[3,1],
              [1,0]])
pmatrix(D, intro=r'D_{2\times 2}=')

B = np.array([[5,1],
              [5,2]])
pmatrix(B, intro=r'B_{2\times 2}=')


    

    






$$C_{2\times 2}=\begin{pmatrix}
  1 & 2\\
  0 & 1\\
\end{pmatrix}$$






    






$$D_{2\times 2}=\begin{pmatrix}
  3 & 1\\
  1 & 0\\
\end{pmatrix}$$






    






$$B_{2\times 2}=\begin{pmatrix}
  5 & 1\\
  5 & 2\\
\end{pmatrix}$$

In [11]:

    

A = np.array([[5,6],
              [3,4]])

pmatrix(rg(A), intro=r'rg(A)=')


    

    






$$rg(A)=\begin{pmatrix}
  2\\
\end{pmatrix}$$

In [12]:

    

pmatrix(inv(C), intro=r'C^{-1}=')

pmatrix(B.T, intro=r'C^{T}=')

pmatrix(B.dot(C), intro=r'BC=')
9
pmatrix(rg(B), intro=r'rg(C)=')

pmatrix(det(B), intro=r'det(C)=')


    

    






$$C^{-1}=\begin{pmatrix}
  1. & -2.\\
  0. & 1.\\
\end{pmatrix}$$






    






$$C^{T}=\begin{pmatrix}
  5 & 5\\
  1 & 2\\
\end{pmatrix}$$






    






$$BC=\begin{pmatrix}
  5 & 11\\
  5 & 12\\
\end{pmatrix}$$






    






$$rg(C)=\begin{pmatrix}
  2\\
\end{pmatrix}$$






    






$$det(C)=\begin{pmatrix}
  5.0\\
\end{pmatrix}$$

In [18]:

    

A = np.array([[2,6],
              [1,3]])
#pmatrix(rg(B), intro=r'rg(B)=')
pmatrix(rg(A), intro=r'rg(A)=')
#pmatrix(rg(A.dot(B)), intro=r'rg(AB)=')


    

    






$$rg(A)=\begin{pmatrix}
  1\\
\end{pmatrix}$$

In [10]:

    

a1 = Symbol('a_{12}')
b1 = Symbol('b_{11}')
c1 = Symbol('c_{22}')
d1 = Symbol('d_{21}')

X =np.array([[a1,b1],
             [c1,d1]])

B = np.array([[5,1],
              [5,2]])

C1 = np.array([[1,1],
               [1,2]])
D1 = np.array([[2,1],
               [1,0]])

C2 = np.array([[1,-1],
               [0,1]])
D2 = np.array([[1,1],
               [0,1]])
pmatrix(B.reshape((4, 1)), intro="X=")


    

    






$$X=\begin{pmatrix}
  5\\
  1\\
  5\\
  2\\
\end{pmatrix}$$

In [4]:

    

pmatrix( (C1.dot(X)).dot(D1))
A = (C1.dot(X)).dot(D1) + (C2.dot(X)).dot(D2)
pmatrix(A)
F = np.array([[3,1,1,1],
              [2,1,0,-1],
              [2,1,5,2],
              [1,0,3,1]])

pmatrix(F, ending=pmatrix(X.reshape((4, 1)),row=True)+"="+pmatrix(B.reshape((4, 1)),row=True))
pmatrix(rg(F), intro=r'rg(F)=')
print("Зничит есть нормальное решение!)")


    

    






$$\begin{pmatrix}
  2*a1 & + & b1 & + & 2*c1 & + & d1 & a1 & + & c1\\
  2*a1 & + & b1 & + & 4*c1 & + & 2*d1 & a1 & + & 2*c1\\
\end{pmatrix}$$






    






$$\begin{pmatrix}
  3*a1 & + & b1 & + & c1 & + & d1 & 2*a1 & + & b1 & - & d1\\
  2*a1 & + & b1 & + & 5*c1 & + & 2*d1 & a1 & + & 3*c1 & + & d1\\
\end{pmatrix}$$






    






$$\begin{pmatrix}
  3 & 1 & 1 & 1\\
  2 & 1 & 0 & -1\\
  2 & 1 & 5 & 2\\
  1 & 0 & 3 & 1\\
\end{pmatrix}\begin{pmatrix}
  a1\\
  b1\\
  c1\\
  d1\\
\end{pmatrix}=\begin{pmatrix}
  5\\
  1\\
  5\\
  2\\
\end{pmatrix}$$






    






$$rg(F)=\begin{pmatrix}
  4\\
\end{pmatrix}$$






    




Зничит есть нормальное решение!)

In [12]:

    

from sympy import Matrix, solve_linear_system
from sympy.abc import a,b,c,d


    

In [14]:

    

system = Matrix(( (3,1,1,1,5), (2,1,0,-1,1), (2,1,5,2,5),(1,0,3,1,2) ))
x = solve_linear_system(system, a,b,c,d)
X =np.array([[x[a],x[b]],[x[c],x[d]] ])


    

In [15]:

    

pmatrix(X,intro="X=")


    

    






$$X=\begin{pmatrix}
  10/11 & 9/11\\
  -2/11 & 18/11\\
\end{pmatrix}$$

In [ ]:

    




    

In [34]:

    

x = Symbol('x')
y = Symbol('y')
pr(integrate(sqrt(4*x-x**2), x))


    

    





$\int \sqrt{- x^{2} + 4 x}\, dx$

In [ ]: