If $A^T$ denotes the transpose of the matrix $A=\left[\begin{array}{lll}0 & 0 & a \\ 0 & b & c \\ d & e &…
If $A^T$ denotes the transpose of the matrix $A=\left[\begin{array}{lll}0 & 0 & a \\ 0 & b & c \\ d & e & f\end{array}\right]$, where $a, b, c, d, e$ and $f$ are integers such that $a b d \neq 0$, then the number of such matrices for which $A^{-1}=A^T$ is
2(3!)
$3(2 !)$
$2^3$
$3^2$
Solution
$
\begin{aligned}
& \text { } A=\left[\begin{array}{lll}
0 & 0 & a \\
0 & b & c \\
d & e & f
\end{array}\right],|A|=-a b d \neq 0 \\
& c_{11}=+(b f-c e), c_{12}=-(-c \mathrm{~d})=c d, \\
& c_{21}=-(-c a)=a e, c_{22}=+(-a d)=-a d, \\
& c_{31}=+(-a b)=-a b, c_{32} \quad c_{23}=-(0)=0 \\
& \operatorname{Adj} A=\left[\begin{array}{ccc}
(b f-c e) & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]
\end{aligned}
$
$A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{a b d}\left[\begin{array}{ccc}b f-c e & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0\end{array}\right]$
$
\begin{aligned}
& A^T=\left[\begin{array}{lll}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right] \\
& \text { Now } A^{-1}=A^T \\
& \Rightarrow \frac{1}{-a b d}\left[\begin{array}{ccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right] \\
& \therefore b f-c e=\left[\begin{array}{ccc}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right]
\end{aligned}
$
$a b d^2=a b, a b^2 d=a d, a^2 b d=b d \quad$...(ii)
abde $=a b c d=a b d f=0$
From (ii),
$\left(a b d^2\right) \cdot\left(a b^2 d\right) \cdot\left(a^2 b d\right)=a b . a d . b d$
$\Rightarrow(a b d)^4-(a b d)^2=0$
$\Rightarrow(a b d)^2\left[(a b d)^2-1\right]=0$
$\because a b d \neq 0, \therefore a b d=\pm 1$
From (iii) and (iv),
$
e=c=f=0
$
From (i) and (v),
$
b f=a e=c d=0
$
From (iv), (v) and (vi), it is clear that $a, b$, $d$ can be any non-zero integer such that $a b d=\pm 1$
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 $=2 \times 2 \times 2=(2)^3$