# 代数|MA20217/MA20219 Algebra 2B代写

Elementary axiomatic theory of rings. Integral domains, fields, characteristic. Subrings and product of rings. Homomorphisms, ideals and quotient rings. Isomorphism theorems. Fields of fractions.

Use Cramer’s rule to solve the system
\begin{aligned} 2 x_{1}-x_{2} &=1 \ 4 x_{1}+4 x_{2} &=20 . \end{aligned}
Solution. The coefficient matrix and right-hand-side vectors are
$$A=\left[\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right] \text { and } \mathbf{b}=\left[\begin{array}{r} 1 \ 20 \end{array}\right]$$
so that
$$\operatorname{det} A=8-(-4)=12$$
and therefore
$$x_{1}=\frac{\left|\begin{array}{rr} 2 & 1 \ 4 & 20 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{36}{12}=3 \quad \text { and } \quad x_{2}=\frac{\left|\begin{array}{rr} 1 & -1 \ 20 & 4 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{24}{12}=2$$

## MA20217/MA20219 COURSE NOTES ：

Since this holds for all $x$, we conclude that $(f+g)+h=f+(g+h)$, which is the associative law for addition of vectors.

Next, if 0 denotes the constant function with value 0 , then for any $f \in V$ we have that for all $0 \leq x \leq 1$,
$$(f+0)(x)=f(x)+0=f(x) .$$
(We don’t write the zero element of this vector space in boldface because it’s customary not to write functions in bold.) Since this is true for all $x$ we have that $f+0=f$, which establishes the additive identity law. Also, we define $(-f)(x)=-(f(x))$ so that for all $0 \leq x \leq 1$,
$$(f+(-f))(x)=f(x)-f(x)=0,$$