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Question

Let A be a 3×3 square matrix and matrices B,C and D are related such that B=adj(A), C=adj(adjA) and D=(adj(adj(adjA))). If det(adj(adj(adj(adj ABCD)))))=|A|k, then the value of k is

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Solution

B=adj(A), C=adj(adj(A)), D=adj(adj(adj(A)))
Let ABCD=E
Then, |adj(adj(adj(adj(E)))|=|E|(31)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

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