# 代数 Algebra MATHS4072_1

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$$\rho\left(J_{p}\right)<\frac{p-1-s}{p-2}$$
then the regions of convergence of the SOR method $\left(\rho\left(\mathscr{L}{\omega}\right)<1\right)$ are For $s=1, \quad \omega \in\left(0, \frac{p}{p-1}\right)$ and for $s=-1, \omega \in\left(\frac{p-2}{p-1}, \frac{2}{1+\rho\left(J{p}\right)}\right)$.

## MATHS4072_1 COURSE NOTES ：

\begin{aligned} &\dot{x}^{k}(t)=G\left(t, x^{k}(t), x^{k-1}(t)\right), t \in[0, T], \ &x^{k}(0)=x_{0} \end{aligned}
for $k=1,2, \ldots .$ Here, the function $x^{k-1}$ is known and $x^{k}$ is to be determined.
Note that the familiar Picard iteration
\begin{aligned} &\dot{x}^{k}(t)=F\left(t, x^{k-1}(t)\right), t \in[0, T], \ &x^{k}(0)=x_{0} \end{aligned}

# 代数 Algebra MAT00010C

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first box. Since the third box is formed from $x_{1}\left(\begin{array}{l}1 \ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \ 1\end{array}\right)=\left(\begin{array}{l}6 \ 8\end{array}\right)$ and $\left(\begin{array}{l}2 \ 1\end{array}\right)$, and the determinant is unchanged by adding $x_{2}$ times the second column to the first column, the size of the third box equals that of the second. We have this.
$$\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|=\left|\begin{array}{ll} x_{1} \cdot 1 & 2 \ x_{1} \cdot 3 & 1 \end{array}\right|=x_{1} \cdot\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|$$

Solving gives the value of one of the variables.
$$x_{1}=\frac{\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|}{\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|}=\frac{-10}{-5}=2$$
The theorem that generalizes this example, Cramer’s Rule, is: if $|A| \neq 0$ then the system $A \vec{x}=\vec{b}$ has the unique solution $x_{i}=\left|B_{i}\right| /|A|$ where the matrix $B_{i}$ is formed from $A$ by replacing column $i$ with the vector $\vec{b}$. Exercise 3 asks for a proof.
For instance, to solve this system for $x_{2}$
$$\left(\begin{array}{ccc} 1 & 0 & 4 \ 2 & 1 & -1 \ 1 & 0 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right)=\left(\begin{array}{c} 2 \ 1 \ -1 \end{array}\right)$$

$$(a+b i)+(c+d i)=(a+c)+(b+d) i$$
\begin{aligned} (a+b i)(c+d i) &=a c+a d i+b c i+b d(-1) \ &=(a c-b d)+(a d+b c) i \end{aligned}