Is there relationship between magnitude of matrix-vector multiplication and determinant of that matrix? If I have a matrix $A$ and vector $x$ is there such a relationship or something similar involving determinants?
$$\|Ax\| \leq |\det A|\|x\|$$
where the absolute values indicate the usual Euclidean norm?
 A: AGTortorella showed that it cannot be true when $det(A) = 0$, but for $det(A) \ne 0$ this inequality also doesn't hold. Consider:
$A=\left(\begin{array}{cc}
1 & a\\ 0 & 1
\end{array}
\right)$
$det(A) = 1$ for every $a$
Let $x = \left(\begin{array}{cc}
1 \\ 1
\end{array}
\right)$
Then:
$Ax = \left(\begin{array}{cc}
1 \cdot 1 + a \cdot 1
\\ 0 \cdot 1 + 1 \cdot 1
\end{array}
\right) = 
\left(\begin{array}{cc}
a + 1
\\ 1
\end{array}
\right)$
So clearly: 
$\|Ax\| \to \infty$ when $a \to \infty$
but
$|det(A)|\|x\|=1 \cdot \sqrt 2$ for every $a$.
A: No.
Because if $\det A=0$ then $|\det A|\|x\|=0,$ for any $x$, while $\|Ax\|$ can take any value varying $x$.
Consider for example
$$A=\left(\begin{array}{cc}
1 & 0\\ 0 & 0
\end{array}
\right).$$
A: If $A$ is an invertible operator of an n dimensional vector space to itself then 
$$\frac{1}{det(A^{-1})^{1/n}} \leq |A|\leq \frac{|A^{-1}|^{n-1}}{det(A^{-1})}$$
This you can get by the relation
$$s_1...s_n = det(A^{-1})$$
where $s_i$ are singular values of $A^{-1}$ which satisfy
$$\frac{1}{|A|}=s_1 \leq ... \leq s_n = |A^{-1}|$$ 
And you get the relation immediately. 
