{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "cb1edcb1",
   "metadata": {},
   "source": [
    "# Systèmes d'équations linéaires"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b802b136",
   "metadata": {},
   "source": [
    "Il est également possible de résoudre des équations linéaires (sous forme polynomiale) ou des systèmes d'équations linéaires"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f3a0b5ba",
   "metadata": {},
   "source": [
    "## Equations polynomiales"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e39a9cae",
   "metadata": {},
   "source": [
    "Pour résoudre une équation du type : $\\sum_{k=0}^{n}(a_k . x^k) = 0$, il est possible d'utiliser une branche particulière de Numpy."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "9707ea3f",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy.polynomial.polynomial as nppol"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1ab66503",
   "metadata": {},
   "source": [
    "On souahite résoudre l'équation : $-6 . x^2 - 2 . x + 4 = 0$ "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b423a358",
   "metadata": {},
   "source": [
    "> **Exécutez les instructions suivantes** et déterminez que contient la variable *X*"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "e8cff478",
   "metadata": {},
   "outputs": [],
   "source": [
    "X = nppol.polyroots( [ 4, -2, -6] )\n",
    "print( X )"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f2d55be2",
   "metadata": {},
   "source": [
    "## Systèmes d'équations linéaires"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "52a60bd9",
   "metadata": {},
   "source": [
    "Soit le système d'équations suivant :\n",
    "\n",
    "$$\\begin{cases} a_1 . x + b_1 . y = c_1 \\\\ a_2 . x + b_2 . y = c_2 \\end{cases}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1f44f79d",
   "metadata": {},
   "source": [
    "Il est possible de le mettre sous une forme matricielle en posant :\n",
    "\n",
    "$$A = \\begin{pmatrix}a_1 & b_1\\\\ a_2 & b_2\\end{pmatrix}  \\text{, }  b = \\begin{pmatrix}c_1\\\\ c_2\\end{pmatrix} \\text{et } X = \\begin{pmatrix}x\\\\y\\end{pmatrix} $$"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "id": "9f54b5ef",
   "metadata": {},
   "source": [
    "Si le système possède une solution, alors la matrice **A** est inversible et le résultat peut s'obtenir par : \n",
    "$$ X = A^{-1} \\cdot b$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "104bbbed",
   "metadata": {},
   "source": [
    "Par exemple, on souhaite résoudre le système suivant :\n",
    "$$\\begin{cases} x + 2 . y = 5 \\\\ 3 . x + 4 . y = 6 \\end{cases}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "55240f04",
   "metadata": {},
   "source": [
    "On définit les matrices suivantes :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "95a89e38",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "A = np.array([[1, 2], [3, 4]])\n",
    "b = np.array([[5], [6]])"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a1e94423",
   "metadata": {},
   "source": [
    "On utilise ensuite la méthode ***solve*** de la bibliothèque **linalg** de **SciPy**."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "5d8b9925",
   "metadata": {},
   "outputs": [],
   "source": [
    "from scipy import linalg\n",
    "X = linalg.solve(A, b)\n",
    "print(X)"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3 (ipykernel)",
   "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": "3.9.16"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}