Condition number of a given linear system $Ax=b$ varying $A$ How would you solve this excersize? I have tried it but I am not posting my calculations since it is wrong and I would spend a lot of time writing it in latex.
The problem.
Find the condition number with respect to $\Vert \cdot \Vert_{\infty}$ of
\begin{equation*}
 \begin{pmatrix}
  1 & d \\ 
  d & 1
 \end{pmatrix}
 \begin{pmatrix}
  x \\
  y
 \end{pmatrix}
 =
 \begin{pmatrix}
  1 \\
  0
 \end{pmatrix}
\end{equation*}
supposing $d \neq \pm 1$.
It should be 
$K_{\infty}(A) = \left\vert\frac{\left\vert d\right\vert + 1}{\left\vert d\right\vert - 1}\right\vert$.
I found two different values depending on $\left\vert d\right\vert < 1$ or $\left\vert d\right\vert > 1$, more specifically if $\left\vert d\right\vert < 1$ I found
\begin{equation*}
 \left\vert \frac{2d}{1-d}\right\vert
\end{equation*}
while if $\left\vert d\right\vert > 1$ I found
\begin{equation*}
 \left\vert \frac{2}{1-d}\right\vert.
\end{equation*}
EDIT
In the two solutions given you use $K(A) = \Vert A^{-1} \Vert \Vert A \Vert$.
My result is different because I used a different definition of $K$ i.e.
$K(d) = \left\Vert G'(d)\right\Vert \frac{\left\Vert d\right\Vert}{\left\Vert G(d)\right\Vert}$ where $G$ is the function $d \mapsto x(d)$ which gives the unique solution depending on the parameter $d$.
In the case in which I want to solve $A x = b$ and $d$ it is contained only in $b$ we have $$x = A^{-1}b$$ so $$G(d) = A^{-1}b \quad \text{and} \quad G'(d) = A^{-1}.$$
We so obtain $$K(d) = \left\Vert A^{-1}\right\Vert \frac{\left\Vert b\right\Vert}{\left\Vert A^{-1}b\right\Vert} = \left\Vert A^{-1}\right\Vert \frac{\left\Vert Ax\right\Vert}{\left\Vert x\right\Vert} \leq \left\Vert A^{-1}\right\Vert \left\Vert A\right\Vert.$$
But I don't understand why in the case in which it is $A$ (not $b$) depending of $d$ we can still use the defintion $K(d) = \left\Vert A^{-1}\right\Vert \left\Vert A\right\Vert$.
 A: I have never delved into this material, but I did some wikipedia-ing.
$$\kappa(A) = ||A^{-1}||\cdot ||A||$$
We first find $A^{-1}$:
$$A^{-1} = \begin{pmatrix} \dfrac{1}{1-d^2} & \dfrac{-d}{1-d^2} \\ \dfrac{-d}{1-d^2} & \dfrac{1}{1-d^2}\end{pmatrix}$$
Since the rows contain the same entries for both $A$ and $A^{-1}$,
$$||A^{-1}|| = \left|\dfrac{1}{1-d^2}\right| + \left|\dfrac{-d}{1-d^2}\right| = \dfrac{1+|d|}{|1-d^2|} = \dfrac{1+|d|}{|(1+|d|)(1-|d|)|} = \dfrac{1}{|1-|d||}$$
and
$$||A|| = \left|1\right| + \left|d\right| = 1+|d|$$
So
$$ \kappa(A) = ||A^{-1}||\cdot ||A|| = \dfrac{1 + |d|}{|1-|d||}$$
A: The relevant definitions are
$$K_{\infty}(A)=||A||_{\infty}||A^{-1}||_{\infty}$$
and
$$||A||_{\infty}=\max_{i=1,2}\sum_{j=1}^2|a_{ij}|$$
Since 
$$A^{-1} = \frac{1}{1-d^2} 
    \begin{pmatrix}
  1 & -d \\ 
  -d & 1
 \end{pmatrix}$$
you have $||A||_{\infty}=1+|d|$ and
$$||A^{-1}||_{\infty} = \frac{1}{|1-d^2|}(1+|d|)=
\frac{1}{|1-|d|^2|}(1+|d|) = 
\frac{1+|d|}{|(1-|d||)(1+|d|)|} = \frac{1}{|1-|d||} 
$$
and therefore 
$$K_{\infty}(A)=||A||_{\infty}||A^{-1}||_{\infty}=\frac{1+|d|}{|1-|d||}$$
