Animated Solution for Mathematics - Matrices and Determinants: Let $S=\left\{A=\left(\begin{array}{ccc}0& 1& c\\ 1& a& d\\ 1& b& e\end{array}\right):a,b,c,d,e\in \{0,1\}\text{ and }|A|\in \{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is ________.

Imagine you are standing before a $3\times 3$ matrix,

where the variables $a,b,c,d,e$ are dancing between the values of $0$ and $1$. It looks like a chaotic puzzle, doesn't it? Five variables, thirty-two possible combinations, and a determinant condition that must land on either $1$ or $-1$.

But here is the secret: mathematics is not about brute force; it is about finding the hidden structure.

Let us begin by expanding the determinant along the first row. Why the first row? Because it contains a zero, which acts as a silent partner, eliminating one of the terms entirely.

When we calculate $|A|$, we get:

This simplifies beautifully to $|A|=c(b-a)-(e-d)$. Suddenly, the chaos begins to subside. We are no longer looking at a matrix; we are looking at a simple algebraic expression.

To make this even clearer, let us define $X=b-a$ and $Y=e-d$. Now, our determinant equation is simply $|A|=cX-Y$.

This is the turning point. We have reduced a complex matrix problem into a simple arithmetic one. But before we celebrate, we must respect the constraints.

Since $a,b,d,e\in \{0,1\}$, the differences $X$ and $Y$ can only take values in the set $\{-1,0,1\}$. This is a crucial realization. If you miss this, you will be lost in a sea of possibilities.

Now, we split the problem into two mutually exclusive cases based on the value of $c$.

If $c=0$, the equation becomes $|A|=-Y$. For the determinant to be $1$ or $-1$, $Y$ must be $1$ or $-1$.

We count the ways to get $Y=1$ (which is $e=1,d=0$) and $Y=-1$ (which is $e=0,d=1$). That gives us two ways for the pair $(d,e)$. Since $a$ and $b$ are independent, they contribute $2\times 2=4$ ways.

Total for Case 1: $2\times 4=8$ matrices.

If $c=1$, the equation becomes $|A|=X-Y$. We need $X-Y$ to be $1$ or $-1$.

For $X-Y=1$, we have $(X,Y)=(1,0)$ or $(0,-1)$. For $X-Y=-1$, we have $(X,Y)=(0,1)$ or $(-1,0)$.

By counting the ways to form each $X$ and $Y$ (remembering that $X=0$ has two ways: $0-0$ and $1-1$), we find that each sub-case yields $4$ ways.

Total for Case 2: $4+4=8$ matrices.

We have navigated the maze. We found $8$ matrices in Case 1 and $8$ matrices in Case 2.

Since these cases are distinct, we simply add them together. The total number of matrices in set $S$ is $8+8=16$.

It is a beautiful, clean result. Remember, the next time you face a complex problem, do not rush. Look for the substitution, define your bounds, and break the problem into logical cases. You have the tools; now go and master the math! The final answer is 16.