] then A2=1 ⇒a2+bc=1ab+bd=0 ac+cd=0bc+d2=1 From these four relations, a2+bc=bc+d2⇒a2=d2 and b(a+d)=0=c(a+d) ⇒a=−d We can take a=1,b=0,c=0,d=−1 as one possible set of values, then A=[
1
0
0
−1
] Clearly A≠I and A≠−I and A=−1 ∴ Statement 1 is true. Also if A≠Itr(A)=0 ∴ Statement 2 is false.