1. Given matrix

We need to count the number of matrices

such that $|A|\in\{-1,1\}.$

So we need the determinant to be $\pm 1$.

2. Compute the determinant

Expand along the first row:

Now,

Hence,

So the condition becomes

3. Case-wise counting based on $c$

Since $c\in\{0,1\}$, split into two cases.

Case 1: $c=0$

Then

Since $d,e\in\{0,1\}$,

We want $d-e=\pm 1$, which happens when:

• $(d,e)=(1,0)$ gives $1$
• $(d,e)=(0,1)$ gives $-1$

So there are $2$ choices for $(d,e)$.

Now $a,b$ are completely free, each in $\{0,1\}$, so there are $2\times 2=4$ choices for $(a,b)$.

Thus total matrices in this case:

Case 2: $c=1$

Then

Let us count the number of $(a,b,d,e)\in\{0,1\}^4$ such that

Rearrange as

Now:

• $b+d$ can be $0,1,2$
• $a+e$ can be $0,1,2$

Let $x=b+d, \qquad y=a+e.$ Then we need $x-y=\pm 1.$

Number of ways to get each sum from two binary variables:

• sum $0$: $1$ way
• sum $1$: $2$ ways
• sum $2$: $1$ way

Now count:

For $x-y=1$

Possible pairs $(x,y)$ are:

• $(1,0)$: $2\cdot 1=2$ ways
• $(2,1)$: $1\cdot 2=2$ ways

Total: $2+2=4.$

For $x-y=-1$

Possible pairs $(x,y)$ are:

• $(0,1)$: $1\cdot 2=2$ ways
• $(1,2)$: $2\cdot 1=2$ ways

Total: $2+2=4.$

Hence total for $c=1$ is $4+4=8.$

4. Total count

Adding both cases,

5. Comparison with stored answer

Our derived answer is $16$, which matches the stored correct answer.