‌As‌A‌adj‌A=|A|I,det(λA)=λn‌det‌A ‌det(3‌adj(−6‌adj(3A)))=33‌det(adj(−6‌adj(3A))) ‌=33(−6‌adj(3A))2 ‌=33(−6)6|3A|4 ‌=3926⋅312⋅(−2)4 ‌=321⋅210 Now comparing with given condition ‌2m+n3mn=210⋅321 ‌m+n=10,mn=21 ‌⇒‌‌m=7,n=3(m>n) ‌∴‌‌4m+2n=28+6=34