Question

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

• A. $2(3!)$
• B. $3(2!)$
• C. $2^3$
• D. $3^2$

Answer 1

Answer: (C)

$A=\left[\begin{array}{ccc}0& 0& a\\ 0& b& c\\ d& e& f\end{array}\right],\left|A\right|=-abd\ne 0$
$c_{11}=+\left(bf-ce\right),c_{12}=-\left(-cd\right)=cd,c_{13}=+\left(-bd\right)=-bd$
$c_{21}=-\left(-ea\right)=ae,c_{22}=+\left(-ad\right)=-ad,c_{23}=-\left(0\right)=0$
$c_{31}=+\left(-ab\right)=-ab,c_{32}=-\left(0\right)=0,c_{33}=0$
$\text{Adj}A=\left[\begin{array}{ccc}\left(bf-ce\right)& ae& -ab\\ cd& -ad& 0\\ -bd& 0& 0\end{array}\right]$
$A^{-1}=\frac{1}{|A|}\left(\text{adj}A\right)=\frac{1}{abd}\left[\begin{array}{ccc}bf-ce& ae& -ab\\ cd& -ad& 0\\ -bd& 0& 0\end{array}\right]$
$A^T=\left[\begin{array}{ccc}0& 0& d\\ 0& b& e\\ a& c& f\end{array}\right]$
Now $A^{-1}=A^T$
$\Rightarrow \frac{1}{-abd}\left[\begin{array}{ccc}bf-ce& ae& -ab\\ cd& -ad& 0\\ -bd& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& d\\ 0& b& e\\ a& c& f\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}bf-ce& ae& -ab\\ cd& -ad& 0\\ -bd& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& -ab{d}^2\\ 0& -a{b}^{2}d& -abde\\ -a^{2}bd& -abcd& -abdf\end{array}\right]$
$\therefore bf-ce=ae=cd=0$ ...(i)
$ab{d}^2=ab,a{b}^{2}d=ad,a^{2}bd=bd$ ...(ii)
$abde=abcd=abdf=0$ ...(iii)
From (ii),
$\left(ab{d}^2\right).\left(a{b}^{2}d\right).\left(a^{2}bd\right)=ab.ad.bd$
$\Rightarrow {\left(abd\right)}^4-{\left(abd\right)}^2=0$
$\Rightarrow {\left(abd\right)}^2\left[{\left(abd\right)}^2-1\right]=0$
$\because abd\ne 0,\therefore abd=\pm 1$ ...(iv) From (iii) and (iv),
$e=c=f=0$ ...(v)
From (i) and (v),
$bf=ae=cd=0$ ...(vi)
From (iv), (v) and (vi), 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 = $2\times 2\times 2=(2{)}^3$