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‎LA_05_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 05\n",
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"\n",
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"# 2 x 2矩阵求逆\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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"这段代码的主要目标是计算一个 $2 \\times 2$ 矩阵的逆矩阵。假设我们有矩阵：\n",
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"\n",
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"$$\n",
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"A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"要计算 $A$ 的逆矩阵 $A^{-1}$，首先需要计算矩阵的 **行列式**（determinant）。行列式是一个标量值，它用于判断矩阵是否可逆，计算公式如下：\n",
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"\n",
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"$$\n",
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"\\det(A) = ad - bc\n",
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"$$\n",
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"\n",
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"如果 $\\det(A) = 0$，那么矩阵是**不可逆的**，即不存在 $A^{-1}$。在代码中，这一步通过 `det = a * d - b * c` 计算，并用 `if det == 0:` 判断是否可逆。如果行列式为零，函数会打印 \"矩阵不可逆，行列式为零\"，并返回 `None`，终止计算。\n",
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"\n",
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"如果矩阵是可逆的，即 $\\det(A) \\neq 0$，则可以按照标准公式计算逆矩阵：\n",
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"\n",
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"$$\n",
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"A^{-1} = \\frac{1}{\\det(A)} \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"在代码中，首先创建矩阵：\n",
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"\n",
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"$$\n",
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"\\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"然后，将其除以 $\\det(A)$ 来得到最终的逆矩阵。这一步在代码里是 `inv_A = np.array([[d, -b], [-c, a]]) / det`。\n",
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"\n",
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"代码示例中，矩阵 $A$ 被定义为：\n",
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"\n",
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"$$\n",
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"A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"其行列式计算如下：\n",
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"\n",
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"$$\n",
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"\\det(A) = (1 \\times 4) - (2 \\times 3) = 4 - 6 = -2\n",
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"$$\n",
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"\n",
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"因此，$A$ 是可逆的，按照逆矩阵公式计算：\n",
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"\n",
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"$$\n",
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"A^{-1} = \\frac{1}{-2} \\begin{bmatrix} 4 & -2 \\\\ -3 & 1 \\end{bmatrix}\n",
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"= \\begin{bmatrix} -2 & 1 \\\\ 1.5 & -0.5 \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"最终，`inverse_2x2_A(A)` 计算并返回了这个逆矩阵。该方法利用 `numpy` 进行矩阵运算，并确保只对可逆矩阵进行计算，从而避免错误。"
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]
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},
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{
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"cell_type": "markdown",
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"id": "b7998c9a-044c-4308-b809-c7458b33a48b",
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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": 4,
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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": "fd8fe07d-9741-4d49-bc75-4292daf3790b",
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"metadata": {},
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"source": [
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"## 自定义函数"
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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": 6,
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"id": "ad0dd3b8-1858-4362-82f0-c17ff891d911",
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"metadata": {},
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"outputs": [],
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"source": [
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"def inverse_2x2_A(A):\n",
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" \"\"\"\n",
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" A = [[a, b],\n",
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" [c, d]]\n",
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" \"\"\"\n",
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" # 提取矩阵元素\n",
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" a, b = A[0]\n",
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" c, d = A[1]\n",
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" \n",
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" # 计算行列式 det(A) = ad - bc\n",
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" det = a * d - b * c\n",
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" \n",
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" # 判断行列式是否为零\n",
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" if det == 0:\n",
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" print(\"矩阵不可逆，行列式为零\")\n",
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" return None\n",
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" \n",
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" # 计算逆矩阵\n",
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" inv_A = np.array([[d, -b], [-c, a]]) / det\n",
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" return inv_A"
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]
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},
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{
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"cell_type": "markdown",
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"id": "5905d558-ffac-4737-868f-7c9b665aeab0",
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"metadata": {},
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"source": [
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"## 定义矩阵"
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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": 8,
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"id": "1e029681-047b-42b1-9f57-6476396e95a8",
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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, 2],\n",
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" [3, 4]])"
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]
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},
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"execution_count": 8,
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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 = [[1, 2], \n",
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" [3, 4]] # 示例矩阵\n",
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"A = np.array(A)\n",
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"A"
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]
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},
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{
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"cell_type": "markdown",
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"id": "50736aaa-bc64-46c9-9527-8dce6b0ba101",
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"metadata": {},
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"source": [
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"## 计算逆矩阵"
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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": 9,
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"id": "72553dc9-98d5-4509-ab4e-679c3668f3b3",
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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. , 1. ],\n",
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" [ 1.5, -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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"inverse_2x2_A(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": null,
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"id": "d86b9119-2028-4797-8c3e-6c7af8fd2995",
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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": "code",
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"execution_count": null,
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"id": "9eac7716-a90e-4c76-ac0f-c81c864dd073",
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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": "code",
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"execution_count": null,
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"id": "3886cb7b-972c-4347-9b6b-f66c8de25cbb",
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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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