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 $abd ≠0$ def then the number of such matrices for which $A^{-1} = A^T$ is:

$2^3$

We have,

$A=\begin{bmatrix}0&0&a\\0&b&c\\d&e&f\end{bmatrix}$

$⇒A^T=\begin{bmatrix}0&0&d\\0&b&e\\a&c&f\end{bmatrix}$ and $A^{-1}=-\frac{1}{abd}\begin{bmatrix}bf-ec&ae&-ab\\cd&-ad&0\\-bd&0&0\end{bmatrix}$

$⇒A^{-1}=\begin{bmatrix}\frac{bf-ec}{abd}&\frac{-e}{abd}&\frac{1}{d}\\\frac{-c}{ab}&\frac{1}{b}&0\\\frac{1}{a}&0&0\end{bmatrix}$

$∴A^T=A^{-1}$

$⇒d^2 =1, b^2 =1, a^2 =1, e =0, c = 0$ and $f = 0$

$⇒a = ±1,b=±1, d=±1, e=0, c = 0$ and $f = 0$

Hence, there are $2^3$ matrices satisfying $A^{-1} = A^T$.