A) 2(3!)
B) 3(2!)
C) 23
D) 32
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \\ \end{matrix} \right],|A|=-abd\ne 0\] \[{{c}_{11}}=+(bf-ce),{{c}_{12}}=-(-cd)=cd,\] \[{{c}_{13}}=+(-bd)=-bd\] \[{{c}_{21}}=-(-ea)=ae,{{c}_{22}}=+(-ad)=-ad,\] \[{{c}_{23}}=-(0)=0\] \[{{c}_{31}}=+(-ab)=-ab,{{c}_{32}}\] \[=-(0)=0,{{c}_{33}}=0\] \[AdjA=\left[ \begin{matrix} (bf-ce) & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \\ \end{matrix} \right]\] \[{{A}^{-1}}=\frac{1}{|A|}(adjA)=\frac{1}{abd}\left[ \begin{matrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \\ \end{matrix} \right]\] \[{{A}^{T}}\left[ \begin{matrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \\ \end{matrix} \right]\]Now\[{{A}^{-1}}={{A}^{T}}\] \[\Rightarrow \]\[\frac{1}{-abd}\left[ \begin{matrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \\ \end{matrix} \right]\]\[=\left[ \begin{matrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \\ \end{matrix} \right]\] \[\Rightarrow \]\[=\left[ \begin{matrix} bf-ce & ae & -ab \\ cd & b-ad & 0 \\ -bd & 0 & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 0 & 0 & -ab{{d}^{2}} \\ 0 & -a{{b}^{2}}d & -abde \\ -{{a}^{2}}bd & -abcd & -abdf \\ \end{matrix} \right]\] \[\therefore \]\[bf-ce=ae=cd=0\] ?(i) \[ab{{d}^{2}}=ab,a{{b}^{2}}d=ad,{{a}^{2}}bd=bd\] ?(ii) \[abde=abcd=abdf=0\] ?(iii) From (ii), \[(ab{{d}^{2}}).(a{{b}^{2}}d).({{a}^{2}}bd)=ab.ad.bd\] \[\Rightarrow \]\[{{(abd)}^{4}}-{{(abd)}^{2}}=0\] \[\Rightarrow \]\[{{(abd)}^{4}}[{{(abd)}^{2}}=1]=0\] \[\because \]\[abd\ne 0,\therefore \] ?(iv) From (iii) and (iv)m\[\] ?(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 \[=2\times 2\times 2={{(2)}^{3}}\]
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