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‎LA_07_01_01.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "f04dd603-a2d1-48ce-8c17-9f1dba8de1ee",
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"metadata": {},
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"source": [
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"Chapter 07\n",
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"\n",
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"# 鸡兔同笼-矩阵乘法\n",
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"《线性代数》 | 鸢尾花书：数学不难"
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]
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},
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{
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"cell_type": "markdown",
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"id": "ccafb456-2453-4c82-8a65-b1963a370cb2",
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"metadata": {},
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"source": [
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"该代码使用线性代数的方法解决经典的“鸡兔同笼”问题。本质上，该问题可以被建模为一个 $2 \\times 2$ 线性方程组，并通过矩阵求逆的方法进行求解。\n",
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"\n",
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"---\n",
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"\n",
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"### **1. 设定线性方程组**\n",
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"设鸡的数量为 $x_1$，兔的数量为 $x_2$，我们得到如下两个方程：\n",
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"$$\n",
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"x_1 + x_2 = 35\n",
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"$$\n",
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"$$\n",
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"2x_1 + 4x_2 = 94\n",
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"$$\n",
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"这个方程组可以用矩阵的形式表示为：\n",
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"$$\n",
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"A \\mathbf{x} = \\mathbf{b}\n",
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"$$\n",
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"其中：\n",
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"$$\n",
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"A = \\begin{bmatrix} 1 & 1 \\\\ 2 & 4 \\end{bmatrix}, \\quad\n",
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"\\mathbf{x} = \\begin{bmatrix} x_1 \\\\ x_2 \\end{bmatrix}, \\quad\n",
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"\\mathbf{b} = \\begin{bmatrix} 35 \\\\ 94 \\end{bmatrix}\n",
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"$$\n",
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"矩阵 $A$ 是系数矩阵，向量 $\\mathbf{x}$ 是未知数，向量 $\\mathbf{b}$ 是右侧的常数列向量。\n",
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"\n",
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"---\n",
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"\n",
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"### **2. 计算矩阵 $A$ 的逆**\n",
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"解方程 $\\mathbf{x} = A^{-1} \\mathbf{b}$ 需要先计算矩阵 $A$ 的逆：\n",
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"$$\n",
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"A^{-1} = \\frac{1}{\\det(A)} \\operatorname{adj}(A)\n",
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"$$\n",
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"行列式 $\\det(A)$ 计算如下：\n",
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"$$\n",
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"\\det(A) = 1 \\cdot 4 - 1 \\cdot 2 = 2\n",
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"$$\n",
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"伴随矩阵（余子式矩阵的转置）：\n",
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"$$\n",
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"\\operatorname{adj}(A) = \\begin{bmatrix} 4 & -1 \\\\ -2 & 1 \\end{bmatrix}\n",
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"$$\n",
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"因此，矩阵 $A$ 的逆为：\n",
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"$$\n",
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"A^{-1} = \\frac{1}{2} \\begin{bmatrix} 4 & -1 \\\\ -2 & 1 \\end{bmatrix}\n",
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"= \\begin{bmatrix} 2 & -0.5 \\\\ -1 & 0.5 \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"---\n",
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"\n",
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"### **3. 验证逆矩阵**\n",
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"代码通过计算 $A^{-1} A$ 和 $A A^{-1}$ 来验证矩阵 $A$ 的逆是否正确：\n",
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"$$\n",
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"A^{-1} A = \\begin{bmatrix} 2 & -0.5 \\\\ -1 & 0.5 \\end{bmatrix}\n",
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"\\begin{bmatrix} 1 & 1 \\\\ 2 & 4 \\end{bmatrix} = \n",
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"\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n",
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"$$\n",
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"$$\n",
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"A A^{-1} = \\begin{bmatrix} 1 & 1 \\\\ 2 & 4 \\end{bmatrix}\n",
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"\\begin{bmatrix} 2 & -0.5 \\\\ -1 & 0.5 \\end{bmatrix} = \n",
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"\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n",
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"$$\n",
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"这表明 $A^{-1}$ 计算正确。\n",
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"\n",
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"---\n",
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"\n",
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"### **4. 计算最终解**\n",
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"最终，使用 $A^{-1} b$ 计算未知数 $\\mathbf{x}$：\n",
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"$$\n",
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"\\mathbf{x} = A^{-1} \\mathbf{b} = \n",
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"\\begin{bmatrix} 2 & -0.5 \\\\ -1 & 0.5 \\end{bmatrix}\n",
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"\\begin{bmatrix} 35 \\\\ 94 \\end{bmatrix}\n",
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"$$\n",
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"进行矩阵乘法：\n",
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"$$\n",
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"x_1 = 2 \\times 35 + (-0.5) \\times 94 = 70 - 47 = 23\n",
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"$$\n",
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"$$\n",
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"x_2 = -1 \\times 35 + 0.5 \\times 94 = -35 + 47 = 12\n",
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"$$\n",
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"因此，鸡的数量是 **23**，兔的数量是 **12**。\n",
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"\n",
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"---\n",
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"\n",
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"### **总结**\n",
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"本代码基于矩阵运算求解 $2 \\times 2$ 线性方程组，核心步骤如下：\n",
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"1. 设定矩阵方程 $A \\mathbf{x} = \\mathbf{b}$。\n",
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"2. 计算 $A$ 的逆矩阵 $A^{-1}$。\n",
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"3. 验证 $A^{-1}$ 是否正确。\n",
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"4. 计算 $A^{-1} b$ 得到未知数的值，即 **鸡 23 只，兔 12 只**。\n",
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"\n",
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"这个方法是利用矩阵求逆来解线性方程组的一种常见方式，但在高维情况下，通常采用高斯消元法或 LU 分解等更高效的数值方法。"
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]
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},
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{
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"cell_type": "markdown",
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"id": "f98f12f8-64c9-4907-9910-7534106881b7",
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"metadata": {},
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"source": [
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"## 初始化"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "d530bc5e-a5b5-41cf-92e4-461541be8fc2",
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"metadata": {},
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"outputs": [],
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"source": [
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"import numpy as np"
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]
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},
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{
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"cell_type": "markdown",
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"id": "dcea75fb-c19a-4f90-a0e9-b8556344174b",
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"metadata": {},
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"source": [
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"## 鸡兔同笼系数矩阵"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "f1044cb0-92c2-4e07-8b42-b5571dfc9101",
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"metadata": {},
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"outputs": [],
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"source": [
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"A = np.array([[1, 1], \n",
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" [2, 4]])\n",
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"# a_1, chicken \n",
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"# a_2, rabbit"
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]
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},
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{
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"cell_type": "markdown",
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"id": "6d759117-5643-4b86-a795-1ebf15aa254b",
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"metadata": {},
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"source": [
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"## 常数列向量"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"id": "954b7525-dcef-47cb-af13-40c741ca3891",
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"metadata": {},
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"outputs": [],
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"source": [
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"b = np.array([[35],\n",
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" [94]])"
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]
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},
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{
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"cell_type": "markdown",
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"id": "7799aec4-4734-49f7-b0c4-2d34344aa7be",
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"metadata": {},
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"source": [
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"## 矩阵A的逆"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 9,
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"id": "aa3dd5d6-6437-4775-9611-64e4fbb5ef65",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([[ 2. , -0.5],\n",
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" [-1. , 0.5]])"
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]
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},
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"execution_count": 9,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"A_inv = np.linalg.inv(A)\n",
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"A_inv"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"id": "dd3bd71d-93b6-480f-bdbd-2b5bc85c8143",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([[1., 0.],\n",
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" [0., 1.]])"
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]
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},
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"execution_count": 11,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# 验证\n",
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"A_inv @ A"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 13,
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"id": "c6cdbfed-8757-4050-b184-844f094ccb31",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([[1., 0.],\n",
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" [0., 1.]])"
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]
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},
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"execution_count": 13,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"A @ A_inv"
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]
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},
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{
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"cell_type": "markdown",
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"id": "eb3be1b7-aad6-4a90-b890-06d7213c4edb",
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"metadata": {},
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"source": [
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"## 求解"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 15,
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"id": "ef4cdee3-8577-47e4-af0f-a2a78aec7401",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"array([[23.],\n",
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" [12.]])"
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]
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},
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"execution_count": 15,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"A_inv @ b"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "03232b1c-9ff6-41a6-8eb9-b5f98c289f77",
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"id": "070c3389-8048-43a3-baa7-6666009bce96",
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"metadata": {},
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"source": [
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"作者\t**生姜DrGinger** \n",
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"脚本\t**生姜DrGinger** \n",
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"视频\t**崔崔CuiCui** \n",
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"开源资源\t[**GitHub**](https://github.com/Visualize-ML) \n",
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"平台\t[**油管**](https://www.youtube.com/@DrGinger_Jiang)\t\t\n",
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"\t\t[**iris小课堂**](https://space.bilibili.com/3546865719052873)\t\t\n",
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"\t\t[**生姜DrGinger**](https://space.bilibili.com/513194466) "
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python [conda env:base] *",
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"language": "python",
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"name": "conda-base-py"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.12.7"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}

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