The given determinant can be expanded to get
$$ |A| = d-e + c(b-a) $$
Since there are five elements to be selected and each of them can be selected in $2$ different ways. So, the total number of possible determinants is $2^5.$ Notice that
$$ |d-e+c(b-a)| \le |d-e|+|c||b-a| \le 1+1 =2. $$
$$ \implies -2 \le |A| \le 2 $$
And since the elements of $A$ are integers, the values that $|A|$ takes are integers. $ \implies |A| \in \lbrace-2,-1,0,1,2\rbrace. $ We claim that there is a bijection between the number of determinants that take even values and those that take odd values. Indeed, if $d=0$ and $|A|$ is odd, then put $d=1$ to get $|A|$ as even. If $d=0$ and $|A|$ is even, then put $d=1$ to get $|A|$ as odd. Therefore, the total number of determinants having an odd value, that is $\lbrace-1,1\rbrace$ is equal to the total number of determinants having an even value, that is $ \lbrace-2,0,2\rbrace$. Since the total number of determinants is $32$, therefore the total number of determinants having an odd value, that is $\lbrace-1,1\rbrace$ is $\frac{32}{2}=16.$
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