We need to find the value of ∣3adj(∣3A∣A2)∣
We know that ∣kA∣=kn∣A∣ where n is the order of the matrix and k is a constant.
⇒∣3A∣=33(2)
⇒∣3adj(∣3A∣A2)∣=33∣adj((33⋅2)A2)∣
We know that adj(kA)=kn−1adj(A)
=33∣(33⋅2)2adj(A2)∣
Now we know that ∣adjA∣=∣A∣n−1
=33((33⋅2)2)3∣adj(A2)∣
=33(33⋅2)6∣A2∣3−1
=33(33⋅2)6(2)4
=210⋅321
Hence, the required answer is 210⋅321=610⋅311