B=adj(A), C=adj(adj(A)), D=adj(adj(adj(A)))
Let ABCD=E
Then, |adj(adj(adj(adj(E)))|=|E|(3−1)4=|E|16
Now, |E|16=|ABCD|16
=|A|16|B|16|C|16|D|16
=|A|16(|A|2)16(|A|4)16(|A|8)16
(∵|B|=|adjA|=|A|2|C|=|adj adjA|=|A|22=|A|4|D|=|adj adj adjA|=|A|23=|A|8)
⇒|E|16=|A|16+32+64+128
⇒|E|16=|A|k=|A|240
∴k=240