{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "# Matrices"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "NB : le module numpy n'est pas présent dans toutes les distributions de Python."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from numpy import array,eye,zeros,ones,random,diag,dot"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A = array([[1, 2],[3, 5]])\n",
    "print(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Matrice unité ou identité :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "I = eye(2)\n",
    "print(I)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "print( 2*A + 3*I)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Remarque :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    " print(2*A + 3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "print(zeros((2,3)))\n",
    "print(diag(array([1,2,3])))\n",
    "print(ones((1,4)))\n",
    "print(random.rand(5))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Addition de matrices"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "B = array([[0,5],[2,-1]])\n",
    "print(A , \" = A\")\n",
    "print(B , \" = B\")\n",
    "print(A+B, \" = A+B\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Produit de matrices"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "print(A , \" = A\")\n",
    "print(B , \" = B\")\n",
    "print(dot(A,B), \" = AxB\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Comparer avec :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "print(A*B)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Inverse"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from numpy import linalg\n",
    "linalg.inv(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "vérification : "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "dot(A,linalg.inv(A))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Puissance de matrice"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "linalg.matrix_power(A,3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Résolution de systèmes linéaires"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Le système $$\\left\\lbrace    \\begin{array}{l} 2x+y=7\\\\3x+4y=8 \\end{array} \\right. $$\n",
    "s'écrit sous forme matricielle : $$AX=C$$\n",
    "avec $A=\\begin{pmatrix}2&1\\\\3&4 \\end{pmatrix}$, $X=\\begin{pmatrix} x\\\\y \\end{pmatrix}$ et $C=\\begin{pmatrix} 7\\\\8 \\end{pmatrix}$.\n",
    "\n",
    "\n",
    "En multipliant (à gauche !) les deux membres de $AX=C$ par $A^{-1}$, on obtient  $$A^{-1} \\times AX=A^{-1} \\times C$$\n",
    "soit \n",
    "$$I_2 X=A^{-1} \\times C$$\n",
    "$$X=A^{-1} \\times C$$\n",
    "\n",
    "Dans notre exemple, $$X=\\begin{pmatrix}0,8&-0,2\\\\-0,6&0,4 \\end{pmatrix} \\times \\begin{pmatrix} 7\\\\8 \\end{pmatrix} = \\begin{pmatrix} 4\\\\-1 \\end{pmatrix}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A = array([[2,1],[3,4]])\n",
    "C = array([[7],[8]])\n",
    "dot(linalg.inv(A) , C)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problèmes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Deux systèmes très semblables :\n",
    "\n",
    "   $$\\left\\lbrace    \\begin{array}{l} x-y=1\\\\x-1.00001*y=0 \\end{array} \\right. $$\n",
    "   et \n",
    "   $$\\left\\lbrace    \\begin{array}{l} x-y=1\\\\x-0.99999*y=0 \\end{array} \\right. $$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mais des solutions très différentes.\n",
    "\n",
    "Solution du premier :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A1 = array([[1,-1],[1,-1.00001]])\n",
    "C1 = array([[1],[0]])\n",
    "dot(linalg.inv(A1) , C1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "solution du second :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A2 = array([[1,-1],[1,-0.99999]])\n",
    "dot(linalg.inv(A2) , C1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.8.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}