Let $$S = \left\{A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\}\right\}$$, where $$|A|$$ denotes the determinant of $$A$$. Then the number of elements in $$S$$ is ________.

The matrix is $$A=\begin{pmatrix}0&1&c\\1&a&d\\1&b&e\end{pmatrix}$$ with $$a,b,c,d,e\in\{0,1\}$$.

First find an explicit expression for the determinant.

Expanding along the first row:
$$|A| \;=\;0\,(a e-d b)\;-\;1\,(1\cdot e-d\cdot 1)\;+\;c\,(1\cdot b-a\cdot 1)$$
$$\Longrightarrow\;|A| \;=\;-(e-d)+c(b-a).$$

Define two simpler variables:
$$x=d-e,\qquad y=b-a,$$
so that
$$|A| \;=\;x+c\,y.$$ Because each entry is either 0 or 1, $$x,\;y\in\{-1,0,1\}.$$ Specifically

• $$x=1$$ when $$(d,e)=(1,0),$$
• $$x=0$$ when $$(d,e)=(0,0)\ \text{or}\ (1,1),$$
• $$x=-1$$ when $$(d,e)=(0,1).$$
• $$y=1$$ when $$(b,a)=(1,0),$$
• $$y=0$$ when $$(b,a)=(0,0)\ \text{or}\ (1,1),$$
• $$y=-1$$ when $$(b,a)=(0,1).$$

The requirement is $$|A|=x+c\,y=\pm1.$$ Treat the two cases for $$c.$$

Case 1: $$c=0$$

Then $$|A|=x.$$ For $$|A|=\pm1$$ we need $$x=\pm1.$$

• $$x=1: (d,e)=(1,0)$$
• $$x=-1: (d,e)=(0,1)$$

That gives 2 choices for $$(d,e).$$

The pair $$(a,b)$$ is unrestricted (4 possibilities).

Total in this case: $$1\;(c)\times2\;(d,e)\times4\;(a,b)=8.$$

Case 2: $$c=1$$

Now $$|A|=x+y.$$ We need $$x+y=\pm1.$$ The admissible ordered pairs $$(x,y)$$ and their counts are

• $$(-1,0)$$: 1 way for $$x$$ × 2 ways for $$y$$ = 2
• $$(0,1)$$: 2 ways for $$x$$ × 1 way for $$y$$ = 2
• $$(0,-1)$$: 2 ways for $$x$$ × 1 way for $$y$$ = 2
• $$(1,0)$$: 1 way for $$x$$ × 2 ways for $$y$$ = 2

Adding these, we have 8 matrices when $$c=1.$$

Hence total in this case: $$1\;(c)\times8=8.$$

Adding the two cases:
$$8+8=16.$$

Therefore, the set $$S$$ contains 16 matrices.