JEE Main 2012 — Mathematics Algebra

$$\begin{aligned} & \text { } A=\left[\begin{array}{lll} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{array}\right],|A|=-a b d \neq 0 \\ & c_{11}=+(b f-c e), c_{12}=-(-c \mathrm{~d})=c d, \\ & c_{21}=-(-c a)=a e, c_{22}=+(-a d)=-a d, \\ & c_{31}=+(-a b)=-a b, c_{32} \quad c_{23}=-(0)=0 \\ & \operatorname{Adj} A=\left[\begin{array}{ccc} (b f-c e) & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0 \end{array}\right] \end{aligned}$$​ A=​00d​0be​acf​​,∣A∣=−abd=0c11​=+(bf−ce),c12​=−(−c d)=cd,c21​=−(−ca)=ae,c22​=+(−ad)=−ad,c31​=+(−ab)=−ab,c32​c23​=−(0)=0AdjA=​(bf−ce)cd−bd​ae−ad0​−ab00​​​ $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{a b d}\left[\begin{array}{ccc}b f-c e & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0\end{array}\right]$A−1=∣A∣1​(adjA)=abd1​​bf−cecd−bd​ae−ad0​−ab00​​ $$\begin{aligned} & A^T=\left[\begin{array}{lll} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{array}\right] \\ & \text { Now } A^{-1}=A^T \\ & \Rightarrow \frac{1}{-a b d}\left[\begin{array}{ccc} b f-c e & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{cccc} b f-c e & a e & -a b \\ c d & -a d & 0 \\ -b d & 0 & 0 \end{array}\right] \\ & \therefore b f-c e=\left[\begin{array}{ccc} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{array}\right] \end{aligned}$$​AT=​00a​0bc​def​​ Now A−1=AT⇒−abd1​​bf−cecd−bd​ae−ad0​−ab00​​⇒​bf−cecd−bd​ae−ad0​−ab00​​∴bf−ce=​00a​0bc​def​​​ $a b d^2=a b, a b^2 d=a d, a^2 b d=b d \quad$abd2=ab,ab2d=ad,a2bd=bd...(ii) abde $=a b c d=a b d f=0$=abcd=abdf=0 From (ii), $\left(a b d^2\right) \cdot\left(a b^2 d\right) \cdot\left(a^2 b d\right)=a b . a d . b d$(abd2)⋅(ab2d)⋅(a2bd)=ab.ad.bd $\Rightarrow(a b d)^4-(a b d)^2=0$⇒(abd)4−(abd)2=0 $\Rightarrow(a b d)^2\left[(a b d)^2-1\right]=0$⇒(abd)2[(abd)2−1]=0 $\because a b d \neq 0, \therefore a b d=\pm 1$∵abd=0,∴abd=±1 From (iii) and (iv), $$e=c=f=0$$e=c=f=0 From (i) and (v), $$b f=a e=c d=0$$bf=ae=cd=0 From (iv), (v) and (vi), it is clear that $a, b$a,b, $d$d can be any non-zero integer such that $a b d=\pm 1$abd=±1 But it is only possible, if $a=b=d=\pm 1$a=b=d=±1 Hence, there are 2 choices for each $a, b$a,b and $d$d. there fore, there are $2 \times 2 \times 2$2×2×2 choices for $a, b$a,b and $d$d. Hence number of required matrices $=2 \times 2 \times 2=(2)^3$=2×2×2=(2)3