{ "cells": [ { "cell_type": "markdown", "id": "73bd968b-d970-4a05-94ef-4e7abf990827", "metadata": {}, "source": [ "Chapter 04\n", "\n", "# 矩阵逆\n", "Book_4《矩阵力量》 | 鸢尾花书:从加减乘除到机器学习 (第二版)" ] }, { "cell_type": "markdown", "id": "17664ff7-4c6b-4539-80c1-07f98b6ef2c2", "metadata": {}, "source": [ "该代码定义了一个 $2 \\times 2$ 矩阵 $A$,并计算其逆矩阵 $A^{-1}$。矩阵 $A$ 为:\n", "\n", "$$\n", "A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix}\n", "$$\n", "\n", "通过求逆操作得到 $A$ 的逆矩阵:\n", "\n", "$$\n", "A^{-1} = \\begin{bmatrix} -2 & 1 \\\\ 1.5 & -0.5 \\end{bmatrix}\n", "$$\n", "\n", "接下来,代码计算矩阵 $A$ 与其逆矩阵 $A^{-1}$ 的乘积,理论上应得到单位矩阵:\n", "\n", "$$\n", "A @ A^{-1} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n", "$$\n", "\n", "这段代码展示了如何使用 `inv` 函数计算矩阵的逆,并验证矩阵与其逆矩阵的乘积结果。" ] }, { "cell_type": "markdown", "id": "1feaec35-bb29-42b9-a011-ad9135b0c638", "metadata": {}, "source": [ "## 导入所需库" ] }, { "cell_type": "code", "execution_count": 1, "id": "5d2a42fe-89e9-4736-b847-a7433aece846", "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", "from numpy.linalg import inv # 导入矩阵求逆函数" ] }, { "cell_type": "markdown", "id": "8ed902be-675e-4438-b550-dbbbf9cf6a66", "metadata": {}, "source": [ "## 定义矩阵A" ] }, { "cell_type": "code", "execution_count": 2, "id": "d8339162-83a4-47b2-9d41-2203ab292820", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[1., 2.],\n", " [3., 4.]])" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A = np.array([[1., 2.], # 定义矩阵A\n", " [3., 4.]])\n", "A" ] }, { "cell_type": "markdown", "id": "fb5a51d4-624e-48fb-b792-54c433bb0df9", "metadata": {}, "source": [ "## 计算矩阵的逆" ] }, { "cell_type": "code", "execution_count": 4, "id": "dee6461e-cf77-4d73-b4a2-552c0bbe4fa7", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[-2. , 1. ],\n", " [ 1.5, -0.5]])" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A_inverse = inv(A) # 计算矩阵A的逆\n", "A_inverse" ] }, { "cell_type": "code", "execution_count": 5, "id": "ff4c814d-c6e7-416a-b027-0a2a28ef8ba1", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[1.00000000e+00, 1.11022302e-16],\n", " [0.00000000e+00, 1.00000000e+00]])" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A_times_A_inv = A @ A_inverse # 计算矩阵A与其逆矩阵的乘积\n", "A_times_A_inv" ] }, { "cell_type": "code", "execution_count": null, "id": "85a80909-2aac-49ed-bb7a-f8cc6b80ee7d", "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "code", "execution_count": null, "id": "ecd322f4-f919-4be2-adc3-69d28ef25e69", "metadata": {}, "outputs": [], "source": [] } ], "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": "3.12.7" } }, "nbformat": 4, "nbformat_minor": 5 }