] ∴bf−ce=ae=cd=0 .......(i) abd2=ab,ab2d=ad,a2bd=bd ......(ii) abde=abcd=abdf=0 .....(iii) From (ii), (abd2).(ab2d).(a2bd)=ab.ad.bd ⇒(abd)4−(abd)2=0 ⇒(abd)2[(abd)2−1]=0 becauseabd≠0,∴abd=±1 ....(iv) From (iii) and (iv), e=c=f=0 ......(v) From (i) and (v), bf=ae=cd=0 ......(iv) From (iv), (v) and (vi), it is clear that a,b,d can be any non-zero integer such that abd=±1 But it is only possible, if a=b=d=±1 Hence, there are 2 choices for each a,b and d. there fore, there are 2×2×2 choices for a,b and d. Hence number of required matrices =2×2×2=(2)3