Find conditioning of the matrix Find conditioning of the following matrix:
$$A=\begin{bmatrix}1& 0\\1&\epsilon\end{bmatrix}.$$
in a $\|.\|_\infty$ norm for $\epsilon > 0$
 A: The condition number of a matrix is given by $\kappa(A) = ||A||||A^{-1}||$, in your case the norm you want to use is $||\cdot||_{\infty}.$
So first you want to find the inverse of this $2\times 2$ matrix. 
$$A^{-1} = \dfrac{1}{\epsilon}\begin{bmatrix} \epsilon & 0 \\ -1 & 1 \end{bmatrix}$$
The $||A||_{\infty} = \max(1+0,1+\epsilon) = \max(1,1+\epsilon) = 1+\epsilon$, since $\epsilon > 0$.
For the $\infty$-norm matrix you sum the absolute value of elements across each row then take the maximum value. 
The $||A^{-1}||_{\infty} = \max(1+0,|-\frac{1}{\epsilon}|+|\frac{1}{\epsilon}|) =  \max(1,\frac{2}{\epsilon}) = \frac{2}{\epsilon}$
So the condition number of your matrix $A$ is given by:
$$\kappa(A) = ||A||_{\infty}||A^{-1}||_{\infty} = (1+\epsilon)\cdot \frac{2}{\epsilon} = \frac{2(1+\epsilon)}{\epsilon}$$
Normally, $\epsilon$ is a very small number $\epsilon<<1$, so this condition number will be very large depending on what value you choose for $\epsilon$. When a condition number is high we call the matrix ill-conditioned. Such matrices are not ideal to work with because the solutions can be fairly inaccurate. 
