这是一份manchester曼切斯特大学ECON10192T作业代写的成功案例

Conversely, if $\boldsymbol{A x} \leqq \mathbf{0}$ then $-\boldsymbol{A} \boldsymbol{x} \in L \cap \mathbb{R}{+}^{m}$. Hence there exists $\xi \geqq \mathbf{0}$ such that $-\boldsymbol{A x}=\boldsymbol{Y} \xi$. Therefore $\boldsymbol{A x}=-\sum{i=1}^{r} \xi_{i} \boldsymbol{y}{i}=\sum{i=1}^{r} \xi_{i}\left(A \boldsymbol{x}{i}\right)$ so that $A\left(x-\sum{i=1}^{r} \xi_{i} x_{i}\right)=0$. It follows that $\left(x-\sum_{i=1}^{r} \xi_{i} x_{i}\right) \in \hat{X}$ and
$$
\boldsymbol{x}-\sum_{i=1}^{r} \xi_{i} \boldsymbol{x}{i}=\sum{i=1}^{s} \mu_{i} \hat{x}{i} \quad \text { for some } \mu{i} \geqq 0 .
$$
Thus
$$
\begin{aligned}
\boldsymbol{x} &=\sum_{i=1}^{r} \xi_{i} \boldsymbol{x}{i}+\sum{i=1}^{3} \mu_{i} \hat{\boldsymbol{x}}_{i} \
&=\boldsymbol{B}\left(\begin{array}{l}
\xi \
\boldsymbol{\mu}
\end{array}\right) \in K(\boldsymbol{B})
\end{aligned}
$$

ECON10192T COURSE NOTES ：

$$
\frac{A_{i j}}{x_{i} x_{j}}=\frac{\tau_{i}}{x_{i}} \text { for any } i, j \in J_{x^{*}}
$$
Similarly,
$$
\frac{A_{r s}}{X_{r} \cdot X_{s}}=\frac{t_{s}}{X_{s}} \text { for any } r, s \in J_{\boldsymbol{x}} .
$$
Since the indices $i$ and $s$ are in $J_{x}$,
$$
A_{\text {is }}=t_{s} x_{i}=\tau_{i} x_{s} \text {, }
$$
which in turn implies that
$$
\frac{\tau_{i}}{x_{i}}=\frac{t_{s}}{x_{s}} .
$$