Find all non-singular $3 \times 3$ matrices, such as $A$ and $A^{-1}$ elements are non-negative Task is to describe all non-singular $3 \times 3$ matrices $A$  for which holds:  all elements of $A$ and $A^{-1}$ is non-negative.
As I discovered linear algebra is  the most problematic part of math for me. Expect to get better with a bit of your help.
I have trouble even approaching the problem, would you provide a slight hint where to go?
 A: The equation $A \cdot A^{-1} = I$ will be key. Also useful is the fact that there is only one way to add non-negative numbers to get zero.
A: Hints:


*

*The inner product of two non-zero vectors from $\Bbb{R}^3$ with non-negative entries is always non-negative. Furthermore it is $=0$ only if the two vectors have no common non-zero components.

*If a row (resp. a column) of $A$ has a non-zero component in position $i$, $i=1,2$ or $3$, then prove that at most one column (resp. a row) of $A^{-1}$ can have a non-zero component in position $i$.


Caveat: It is not necessary that row number $i$ has its (only) non-zero entry on column $i$.
A: We know that $A A^{-1}= I$.
 Then suppose $X=A^{-1}$ => Elements of X could be derived from following set of linear systems:
$\begin{pmatrix}
    a & 0 & 0 & | & 1\\ 
    a_{21} &a_{22} &a_{23} & | & 0\\
    a_{31} &a_{32} &a_{33} & | & 0\\ 
    \end{pmatrix} => x_{1,1}= \frac 1 a$
$  \begin{pmatrix}
    a & 0 & 0 & | & 0\\ 
    a_{21} &a_{22} &a_{23} & | & 1\\
    a_{31} &a_{32} &a_{33} & | & 0\\ 
    \end{pmatrix} => x_{2,1}= 0$
$\begin{pmatrix}
    a & 0 & 0 & | & 0\\ 
    a_{21} &a_{22} &a_{23} & | & 0\\
    a_{31} &a_{32} &a_{33} & | & 1\\ 
    \end{pmatrix} => x_{3,1}= 0$
This derivations prove second bullet from Jyrki's answer.
Also, in order to satisfy requirements, we need: $|A| > 0$,
$Minor_{i,j} \ge 0 $ if $i+j = 0 (mod 2)$
$Minor_{i,j} \le 0 $ if $i+j = 1 (mod 2)$
Then, A is:
$\begin{pmatrix}
    a & 0 & 0\\ 
    0 &b &0\\
   0 &0 &c\\
    \end{pmatrix}, a,b,c > 0$
