Conditions for the value of a determinant to be zero The theory states that the value of a determinant will be zero if it contains a row or column full of zeroes or if it has two identical rows or two rows proportional to each other. 

Similarly, can we say that the value of the determinant is zero only if it satisfies the above mentioned conditions?

Could someone please explain me this?
 A: Nobody can explain it since it is not true. For instance,$$\begin{vmatrix}0&1&1\\1&0&1\\1&1&2\end{vmatrix}=0,$$in spite of the fact that none of the conditions that you mentioned hold. The general rule is: a determinant is $0$ if and only if some row is a linear combination of the other rows (and, of course, this is also equivalent to the assertion that some column is a linear combination of the other columns).
A: At first, some definitions: 

A function $f\colon X^n\to\mathbb{R}$ is multilinear when is linear respect all their variables. 

and other

A function $f\colon X^n\to\mathbb{R}$ is alternating when:
   $$ f(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)=-f(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_n) $$

At least 

A determinant function $f\colon X^n\to\mathbb{R}$ is a multilinear alternating function.

So if we have $x_i=x_k$ for some $i,k\leq n$:
$$ f(x_1,\ldots,x_i,\ldots,x_k,\ldots,x_n)=-f(x_1,\ldots,x_k,\ldots,x_i,\ldots,x_n) $$
$$ =-f(x_1,\ldots,x_i,\ldots,x_k,\ldots,x_n) \mbox{ since }x_i=x_k$$
so $$f(x)=-f(x) \Leftrightarrow f(x)=0, \mbox{ where } x=(x_1,\ldots,x_n)$$
So, If $x$ has two equal elements, $f(x)=0$.
If $x$ has some $0$: 
$$f(x_1,\ldots,0,\ldots,x_n)=f(x_1,\ldots,0+0,\ldots,x_n)$$
$$f(x_1,\ldots,0,\ldots,x_n)=f(x_1,\ldots,0,\ldots,x_n)+f(x_1,\ldots,0,\ldots,x_n)\mbox{ since } f \mbox{ is linear }$$
$$f(x)=f(x)+f(x)$$
$$f(x)=0$$
