MATH1131 - 矩阵基础与线性方程组
矩阵运算、行列式、逆矩阵与线性系统求解
一个 3×2 矩阵:
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$$其中 $a_{11} = 1$, $a_{12} = 2$, $a_{21} = 3$, 等等。
矩阵加法: $(A + B)_{ij} = a_{ij} + b_{ij}$
标量乘法: $(kA)_{ij} = k \cdot a_{ij}$
乘积元素公式:
$$(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$$即:第 i 行与第 j 列的点积
给定:
$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$计算 AB:
$$AB = \begin{bmatrix} 1×5+2×7 & 1×6+2×8 \\ 3×5+4×7 & 3×6+4×8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$$2×2 行列式:
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$3×3 行列式(按第一行展开):
$$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}$$计算:
$$\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix}$$解:按第一行展开
$$= 1\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}$$ $$= 1(0-24) - 2(0-20) + 3(0-5)$$ $$= -24 + 40 - 15 = 1$$对于矩阵 $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
$$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$前提:det(A) = ad - bc ≠ 0
求矩阵的逆:$A = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$
步骤:
线性方程组可以写成:Ax = b
方程组:
$$\begin{cases} 2x + 3y = 7 \\ x - y = 1 \end{cases}$$矩阵形式:
$$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$$解方程组:
$$\begin{cases} 3x + y = 11 \\ 2x + 4y = 18 \end{cases}$$步骤:
答案: x = 2.6, y = 3.2
先旋转45°,再放大2倍:
$$T = S(2) \cdot R(45°) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}$$ $$= \begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ \sqrt{2} & \sqrt{2} \end{bmatrix}$$MATH1131 - Matrix Fundamentals and Linear Systems
Matrix Operations, Determinants, Inverse Matrices and Linear System Solutions
Part 1: Matrix Basics and Operations
Part 2: Determinants and Inverse Matrices
Product Element Formula:
$$(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$$i.e., dot product of row i and column j
For matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
$$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$Provided: det(A) = ad - bc ≠ 0
Linear systems can be written as: Ax = b