Finding vector which have a certain norm. 
Let $$A=\begin{pmatrix}
9 & 5\\
11 & 6
\end{pmatrix}$$
Find the conditionnumber $K(A)$ (for $\|\cdot \|_\infty$), and then try to find $b, r \in \mathbb{R}^2$ such that (for $Ax= b$ and $Ae = r$) with $\|r\|_\infty = 2^{-24}$: 
$$\dfrac{\|e\|_\infty \cdot \|b\|_\infty}{\|r\|_\infty \cdot \|x\|_\infty} = K(A)$$ 

I have calculated $K(A)$ to be 340.
Let $r = \begin{pmatrix}2^{-24}\\0 \end{pmatrix}$, then $e = A^{-1} \cdot r$ results in $\begin{pmatrix}-6\cdot 2^{-24}\\11\cdot 2^{-24} \end{pmatrix}$ or $\|e\|_\infty = 11\cdot 2^{-24}$.
Then the fraction is reduced to:
$$ 11 \cdot \|b\|_\infty = 340 \cdot \|x\|_\infty$$ wich translates into $\|b\|_\infty \approx 31 \|x\|_\infty$
So I need to find an $x$ such that the product with $A$ makes it about $31$ times as big. But how can I find such values. I tried some trail and error method, but I don't seem to get there.
 A: Hereafter all norm are considered to be infinite norms. Consider two systems 
$$
Ax = b\\
A(x+e) = b + r
$$
The second one can be thought as a perturbation of the first one. Then $\frac{||e||}{||x||}$ can be thought as relative error of the solution and $\frac{||r||}{||b||}$ as a relative right hand side perturbation.
Let's make some upper estimates on the solution relative error:
$$
\frac{||e||}{||x||} = \frac{||A^{-1}r||}{||x||}
\overset{(a)}{\leq} \frac{||A^{-1}||\cdot ||r||}{||x||} =
\frac{||A^{-1}||\cdot ||r|| \cdot||b||}{||x||\cdot||b||} =
\frac{||A^{-1}||\cdot ||r|| \cdot||Ax||}{||b||\cdot||x||} \overset{(b)}{\leq}
||A^{-1}||\cdot||A||\cdot \frac{||r||}{||b||}.
$$
Here we twice used norm property
$$
||Ax|| \leq ||A||\cdot||x||.
$$
Note that the property is tight, that means we can find such vector $x_0 \neq 0$ satisfying $||Ax_0|| = ||A|| \cdot ||x_0||$.
For infinity norm
$$
||A|| = \max_i \sum_j |a_{ij}| = \sum_j |a_{i_0 j}|\\
(x_0)_j = \operatorname{sign} a_{i_0 j}
$$
where $i_0$ is the row where the matrix norm is achieved.
Proof. Note that $||x_0|| = 1$. Forward: 
$$
z = Ax_0 = \begin{pmatrix}\vdots\\\sum_j a_{i_0j}\operatorname{sign} a_{i_0 j} \\
\vdots\end{pmatrix} = 
\begin{pmatrix}\vdots\\\sum_j |a_{i_0 j}| \\
\vdots\end{pmatrix}
$$
Thus $||z|| \leq \max_i \sum_j |a_{ij}|$.
Backward. Let $y = Ax$
$$
|y_i| = |\sum_j a_{ij} x_j| \leq \sum_j |a_{ij}| |x_j| \leq  ||x||\sum_j |a_{ij}|
$$
so
$$
||y|| = \max_i |y_i| \leq  ||x||\max_i\sum_j |a_{ij}|
$$
Since $||z||$ can not exceed $\max_i\sum_j |a_{ij}|$ they are exactly equal each other. We've proved that 
$$
||A|| = \max_i \sum_j |a_{ij}|
$$
and found the extremal vector $x_0$. That vector is defined up to multiplying by an arbitrary constant.
Back to the problem. We used the matrix norm property twice, noted as (a) and (b).
(a) $
||A^{-1} r|| \leq ||A^{-1}||\cdot ||r||.
$
For this to be equality we can choose $r$ as $\epsilon x_0$ for matrix $A^{-1}$. Since $A^{-1}$ is
$$
A^{-1} = \begin{pmatrix}
-6&5\\11&-9
\end{pmatrix},
$$
its norm is $20$ and is achieved in second row on vector $x_0 = (1, \; -1)^\mathsf{T}$. So $r = (\epsilon, \; -\epsilon)^\mathsf{T}$
(b) $
||A x|| \leq ||A ||\cdot ||x||.
$. For $A$ 
$$
A = \begin{pmatrix}
9&5\\11&6
\end{pmatrix},
$$
norm is $17$ achieved in second row on vector $x_0 = (1,\;1)^\mathsf{T}$. Let's take it as $x$ as it is. Corresponding $b$ is $b = (14,\;17)^\mathsf{T}$.
A: Note that $\|A\|_\infty = 17, \|A^{-1}\|_\infty = 20$. And $\operatorname{cond}_\infty A = 17 \cdot 20$.
Let $x_1 = (1,1)^T, x_2 = (1,-1)^T$, note both have unit norm.
$\|A x_1 \|_\infty = 17$, $\|A^{-1} x_2 \|_\infty = 20$, and so
${\|A^{-1} x_2 \|_\infty \over \|x_2\|_\infty } {\|A x_1 \|_\infty \over \|x_1\|} = 340 $.
