| Solving the determinant, we have |P|=1(12−12)−α(4−6)+3(4−6) ⇒|P|=0−(−2α)+3(−2) |P|=2α−6.....(1) ∵P=adj(A) (Given) ∴|P|=|adj(A)| ⇒|P|=|A|(3−1)=42(∵|adj(A)|=|A|(n−1); whereas n is the order of matrix A) |P|=16.....(2) From eq n(1)&(2), we have 2α−6=16 2α=22 ⇒α=11