If $$A^T$$ denotes the transpose of the matrix $$A = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix}$$, where $$a, b, c, d, e$$ and $$f$$ are integers such that $$abd \neq 0$$, then the number of such matrices for which $$A^{-1} = A^T$$ is

Condition $$A^{-1}=A^{T}$$ means that $$A$$ is an orthogonal matrix, i.e. $$AA^{T}=A^{T}A=I_{3}$$. In an orthogonal matrix, every row (and every column) is a unit vector that is orthogonal to the other two rows (or columns).

Because all entries are integers, a unit vector can contain only $$0,\,1,\,\! -1$$; moreover exactly one entry must be $$\pm1$$ and the others must be $$0$$. (Otherwise its squared length would exceed $$1$$.) Thus every row of $$A$$ is a “signed” standard basis vector.

Write the given matrix as $$A=\begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix},$$ with $$a,b,d\neq0$$. Since the non-zero entries can only be $$\pm1$$ we already have $$a,b,d\in\{\pm1\}$$.

Step 1: Orthogonality of row 1 and row 2
Row 1 is $$[\,0,\;0,\;a\,]$$, Row 2 is $$[\,0,\;b,\;c\,]$$. Their dot product must vanish: $$0\cdot0+0\cdot b+a\cdot c=a\,c=0\;$$, so $$c=0$$.

Step 2: Orthogonality of row 1 and row 3
Row 3 is $$[\,d,\;e,\;f\,]$$. Dotting with Row 1 gives $$0\cdot d+0\cdot e+a\cdot f=a\,f=0\;$$, hence $$f=0$$.

Step 3: Orthogonality of row 2 and row 3
Row 2 is now $$[\,0,\;b,\;0\,]$$. Their dot product is $$0\cdot d+b\cdot e+0\cdot0=b\,e=0\;$$, so $$e=0$$.

Step 4: Unit-length of each row
Row 1 already has squared length $$a^{2}=1$$ (since $$a=\pm1$$).
Row 2 has squared length $$b^{2}=1$$ (since $$b=\pm1$$).
Row 3 has squared length $$d^{2}=1$$ (since $$d=\pm1$$).
Thus all rows are unit vectors and pairwise orthogonal, satisfying orthogonality completely.

After these deductions the matrix must be

$$A=\begin{bmatrix} 0 & 0 & a \\[4pt] 0 & b & 0 \\[4pt] d & 0 & 0 \end{bmatrix}, \qquad\text{where } a,b,d\in\{\pm1\}.$$

Step 5: Counting the matrices
Each of $$a,\,b,\,d$$ can independently take two values ($$+1$$ or $$-1$$). Hence the total number of admissible matrices is $$2\times2\times2=2^{3}=8$$.

Therefore the required number of matrices is $$2^{3} = 8$$.

Option C which is: $$2^{3}$$