Mathematics - Matrix Question with Solution | TestHub

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