${c_{11}} = + \left( {bf - ce} \right),{c_{12}} = - \left( { - cd} \right) = cd,{c_{13}} = + \left( { - bd} \right) = - bd$
${c_{21}} = - \left( { - ea} \right) = ae,{c_{22}} = + \left( { - ad} \right) = - ad,{c_{23}} = \left( { - 0} \right) = 0$
${c_{31}} = + \left( { - ab} \right) = - ab,{c_{32}} = - \left( 0 \right) = 0,{c_{33}} = 0$
Adj $A = \left[ {\begin{array}{*{20}{c}}
{\left( {bf - ce} \right)}&{ae}&{ - ab}\\
{cd}&{ - ad}&0\\
{ - bd}&0&0
\end{array}} \right]$
${A^{ - 1}} = \frac{1}{{\left| A \right|}}\left( {adj\,A} \right) = \frac{1}{{abd}}\left[ {\begin{array}{*{20}{c}}
{bf - ce}&{ae}&{ - ab}\\
{cd}&{ - ad}&0\\
{ - bd}&0&0
\end{array}} \right]$
${A^T} = \left[ {\begin{array}{*{20}{c}}
0&0&d\\
0&b&e\\
a&c&f
\end{array}} \right]$
Now ${A^{ - 1}} = {A^T}$
$ \Rightarrow \frac{1}{{ - abd}}\left[ {\begin{array}{*{20}{c}}
{bf - ce}&{ae}&{ - ab}\\
{cd}&{ - ad}&0\\
{ - bd}&0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&d\\
0&b&e\\
a&c&f
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{bf - ce}&{ae}&{ - ab}\\
{cd}&{ - ad}&0\\
{ - bd}&0&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&{ - ab{d^2}}\\
0&{ - a{b^2}d}&{ - abde}\\
{ - {a^2}bd}&{ - abcd}&{ - abdf}
\end{array}} \right]$
$\therefore bf - ce = ae = cd = 0\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$
$ab{d^2} = ab,a{b^2}d = ad,{a^2}bd = bd\,\,.....\left( {ii} \right)$
$abde = abcd = abdf = 0\,\,\,\,\,\,....\left( {iii} \right)$
From $(ii)$,
$\left( {ab{d^2}} \right).\left( {a{b^2}d} \right).\left( {{a^2}bd} \right) = ab.ad.bd$
$ \Rightarrow {\left( {abd} \right)^4} - {\left( {abd} \right)^2} = 0$
$ \Rightarrow {\left( {abd} \right)^2}\left[ {{{\left( {abd} \right)}^2} - 1} \right] = 0$
$\because $ $abd \ne 0,\,\,\,\,\,\,\,\therefore \,\,\,\,\,\boxed{abd = \pm 1\,}\,\,\,\,\,\,\,\,....\left( {iv} \right)$
From $(iii)$ and $(iv)$
$\boxed{e = c = f = 0}\,\,\,\,\,\,....\left( v \right)$
from $(i)$ and $(v)$,
$bf = ae = cd = 0\,\,\,\,\,\,....\left( {vi} \right)$
From $(iv),(v)$ and $(iv)$, it is clear that $a,b,d$ can be any non-zero integer such that ${abd = \pm 1}$
But it is only possible, if $a = b = d = \pm 1$
Hence, there are $2$ choices for each $a,b$ and $d$. there fore, there are $2 \times 2 \times 2$ choices for $a,b$ and $d$. Hence number of required matrices $ = 2 \times 2 \times 2 = {\left( 2 \right)^3}$
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