Notebook

TestDemo¶

Author: Marcos Bujosa

In [1]:

from nacal import *

Repositorio de la librería en GitHub
Repositorio del proyecto en Pypi
Repositorio con notebooks de jupyter en GitHub
Repositorio del libro en GitHub

$ \newcommand{\Rr}{{\mathbb{R}}} \newcommand{\Zz}{{\mathbb{Z}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\getItem}{\pmb{\mid}} \newcommand{\getitemR}[1]{\getItem{#1}} \newcommand{\getitemL}[1]{{#1}\getItem} \newcommand{\elemR}[2]{{#1}^{\phantom{\T}}_{\getitemR{#2}}} \newcommand{\elemRP}[2]{{\big(#1\big)}^{\phantom{\T}}_{\getitemR{#2}}} \newcommand{\elemRPE}[2]{\big({#1}^{\phantom{\T}}_{\getitemR{#2}}\big)} \newcommand{\elemL}[2]{ {_{\getitemL{#2}}#1} } \newcommand{\elemLP}[2]{ {_{\getitemL{#2}}\left(#1\right)} } \newcommand{\elemLPE}[2]{\left( {_{\getitemL{#2}}#1} \right)} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\Vect}[2][{}]{{\boldsymbol{#2}}_{#1}} \newcommand{\eleVR}[2] {\elemR {\Vect{#1}}{#2}} % con subindices \newcommand{\eleVRP}[2] {\elemRP {\Vect{#1}}{#2}} % con subindices y paréntesis interior \newcommand{\eleVRPE}[2]{\elemRPE{\Vect{#1}}{#2}} % con subindices y paréntesis exterior %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\VectC}[2][{}] {\elemR {\Mat{#2}}{#1}} % con subindices \newcommand{\VectCP}[2][{}] {\elemRP {\Mat{#2}}{#1}} % con subindices y paréntesis \newcommand{\VectCPE}[2][{}]{\elemRPE{\Mat{#1}}{#2}} % con subindices y paréntesis exterior \newcommand{\VectF}[2] {\elemL{\Mat{#1}}{#2}} % con subindices \newcommand{\VectFP}[2] {\elemL{(\Mat{#1})}{#2}} % con subindices y paréntesis \newcommand{\VectFPE}[2] {\elemLPE{\Mat{#1}}{#2}} % con subindices y paréntesis exterior %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\mat}[1]{\boldsymbol{\mathsf{#1}}} \newcommand{\Mat} [2][{}]{{\mat{#2}}_{#1}} \newcommand{\T}{\intercal} \newcommand{\MatT}[2][{}]{{\mat{#2}}^{\T}_{#1}} \newcommand{\VectCC}[2][{}] {\elemRR {\Mat{#2}}{#1}} % con () \newcommand{\VectCCC}[2][{}] {\elemRRR{\Mat{#2}}{#1}} % con texto "col" %SELECCIÓNA de FILAS y COlUMNAS DE UNA MATRIZ TRANSPUESTA PARA GENERAR UN VECTOR DE Rn \newcommand{\VectTC}[2][{}] {\elemR{\MatT{#2}\!}{#1}} % con subindices \newcommand{\VectTCC}[2][{}] {\elemRR{ \MatT{#2}}{#1}} % con () \newcommand{\VectTCCC}[2][{}] {\elemRRR{\MatT{#2}}{#1}} % con texto "col" \newcommand{\dotprod}[2][{}] {\Vect{#1}\cdot\Vect{#2}} $

Matrices¶

Definimos una matriz (en esta ocasión fila a fila)

In [2]:

A = Matrix([[2,-1,0],[-1,2,-1],[0,-1,2]])
A

Out[2]:

$\left[ \begin{array}{ccc}2&-1&0\\-1&2&-1\\0&-1&2\\ \end{array} \right]$

Calculemos su inversa mediante eliminación por columnas de izquierda a derecha. Para ello pongamos la matriz identidad de orden 3 a su derecha

In [3]:

Aampliada = A.concatena(I(3),1)
Aampliada

Out[3]:

$\left[ \begin{array}{ccc|ccc}2&-1&0&1&0&0\\-1&2&-1&0&1&0\\0&-1&2&0&0&1\\ \end{array} \right]$

Y ahora apliquemos la elimnación

In [5]:

A.inversa(1)

$\displaystyle \left[ \begin{array}{ccc}2&-1&0\\-1&2&-1\\0&-1&2\\ \hline 1&0&0\\0&1&0\\0&0&1\\ \end{array} \right]\xrightarrow{\underset{\begin{subarray}{c} \left[\left(2\right)\mathbf{2}\right]\\\left[\left(1\right)\mathbf{1}+\mathbf{2} \right]\end{subarray}}{\pmb{\tau}}}\left[ \begin{array}{ccc}2&0&0\\-1&3&-1\\0&-2&2\\ \hline 1&1&0\\0&2&0\\0&0&1\\ \end{array} \right]\xrightarrow{\underset{\begin{subarray}{c} \left[\left(3\right)\mathbf{1}\right]\\\left[\left(1\right)\mathbf{2}+\mathbf{1} \right]\\\left[\left(3\right)\mathbf{3}\right]\\\left[\left(1\right)\mathbf{2}+\mathbf{3} \right]\end{subarray}}{\pmb{\tau}}}\left[ \begin{array}{ccc}6&0&0\\0&3&0\\-2&-2&4\\ \hline 4&1&1\\2&2&2\\0&0&3\\ \end{array} \right]\xrightarrow{\underset{\begin{subarray}{c} \left[\left(2\right)\mathbf{1}\right]\\\left[\left(1\right)\mathbf{3}+\mathbf{1} \right]\\\left[\left(2\right)\mathbf{2}\right]\\\left[\left(1\right)\mathbf{3}+\mathbf{2} \right]\end{subarray}}{\pmb{\tau}}}\left[ \begin{array}{ccc}12&0&0\\0&6&0\\0&0&4\\ \hline 9&3&1\\6&6&2\\3&3&3\\ \end{array} \right]\xrightarrow{\underset{\begin{subarray}{c} \left[\left(\frac{1}{12}\right)\mathbf{1}\right]\\\left[\left(\frac{1}{6}\right)\mathbf{2}\right]\\\left[\left(\frac{1}{4}\right)\mathbf{3}\right]\end{subarray}}{\pmb{\tau}}}\left[ \begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\\ \hline \frac{3}{4}&\frac{1}{2}&\frac{1}{4}\\\frac{1}{2}&1&\frac{1}{2}\\\frac{1}{4}&\frac{1}{2}&\frac{3}{4}\\ \end{array} \right]$

Out[5]:

$\left[ \begin{array}{ccc}\frac{3}{4}&\frac{1}{2}&\frac{1}{4}\\\frac{1}{2}&1&\frac{1}{2}\\\frac{1}{4}&\frac{1}{2}&\frac{3}{4}\\ \end{array} \right]$

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