The correct option is D |3Adj(2A)|=1728
Since, A2=I
⇒A=A−1
Thus A is an involutory and diagonal matrix,
Let A=⎡⎢⎣a000b000c⎤⎥⎦ (a,b,c≠0).
Given, A2=I
⇒|A|2=1⇒|A|=−1
{∵ entries are non-positive}
Thus |A|=−1
As |A|≠0,A−1 exists
|2A|=−8|3Adj(2A)|=27|Adj(2A)|=27(−8)2=1728