Hi,

Here are two questions:

1) I am given a square matrix $A$. Its determinant does not equal $0$.
Let's compute its condition number, according to the infinity norm,
i.e.
\[
\kappa(A)=||A||_{\infty}\cdot ||A^{-1}||_{\infty}.
\]
Elements of $A$, i.e. $A_{i,j}$, $1\leq i\leq n$, $1\leq j\leq n$,
where $A$ is of dimension $n$, are known with an infinite precision,
but are simple digits (i.e. numbers in the set $\{1, \cdots, 9\}$).
There is thus one significative digit for every $A_{i,j}$. As $\log(1)
=0$, can we say that $A$ is ill-conditioned when we know nothing but
the mere fact that
\[
\kappa(A)>1?
\]

2) I am given a square matrix $A$. Its determinant equals $0$. Can we
say that $A$ is ill-conditioned? The system $Ax=b$ has actually a
countable infinite number of solutions (because of the free variables,
i.e. parameters), but can we still say that it is ill-conditioned? If
we set all the parameters to a fixed value, will it be influenced by a
precision error?

With some less rigorous ideas, $\det A =0$ implies that $A^{-1}$ is
infinite. Consequently, if $A^{-1}$ is infinite, its infinite norm is
infinite, and the $Ax=b$ system is thus ill-conditioned. Am I right?