Let A be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$. Let $A$ be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$. Then we can conclude that 
(a)  $|A| > 0 (|A|$ denotes the determinant of A).
(b) $A$ is a positive definite matrix.
(c)  $B = A^2$ is a positive definite matrix.
(d) $C = A^{-1}$ is a matrix with entries in $(0,\infty)$.
I took an example matrix $A=
        \begin{bmatrix}
        1 & 2\\
        2 & 1 \\
        \end{bmatrix}
$, and saw that options (a), (b) and (d) are wrong. 
But what is the proper approach for this problem ?
 A: You are right about (a), (b), (d).
Eigen values of $A^2$ can not be negative.. 
Again , eigen values of $A$ may not be $0$ , as $A$ is non-singular. So , eigen values of $A^2$ can never zero. Thus , all eigen values of $A^2$ are positive.
So, $A^2$ is positive definite. So, (c) is TRUE.
A: You are right about a,b,d. Here is another approach to c, which does not rely on eigenvalues.
Let $A$ be non-singular and symmetric. Then $B=A^2 = A^TA$ is positive definite.
Take a vector $x\ne0$, then it holds
$$
x^TBx = x^TA^TAx=(Ax)^T(Ax) = \|Ax\|^2 >0,
$$
where we used invertibility of $A$: since $x\ne 0$, $Ax$ cannot be zero. This proves positive definitness of $B$.

Edit: this result is true for symmetric, non-singular, real matrix $A$, the entries do not need to be positive. It is also true if $A\in \mathbb C^{n,n}$ is Hermitian, non-singular. However, the proof would be more complex.
A: For a, b and d your approach is right you have a counterexample. Let's take a look at c. $A$ symmetric non singular and therefore is diagonalisable with real nonzero eigenvalues. $A^2$ will then have positive eigenvalues (the squares of those of $A$) on top of being symmetric so it is positive definite. Non singular is an essential assumption as $\begin{bmatrix} 1&1\\1&1\end{bmatrix}$ shows.
