Book4_Power-of-Matrix/Book4_Ch04_Python_Codes/Bk4_Ch04_11.ipynb

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    "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)"
   ]
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  {
   "cell_type": "markdown",
   "id": "17664ff7-4c6b-4539-80c1-07f98b6ef2c2",
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   "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",
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   "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": []
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   "cell_type": "code",
   "execution_count": null,
   "id": "ecd322f4-f919-4be2-adc3-69d28ef25e69",
   "metadata": {},
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   "source": []
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