Given, __A=5a−b32 and A adj A=AAT Clearly __A(adj⁡A)=A∣I2∵ if A is square matrix of order n, then A(adj⁡A)=(adj⁡A)⋅A=A∣InA(adj⁡A)=5a−b32I2A(adj⁡A)=(10a+3b)I2A(adj⁡A)=(10a+3b)1001A(adj⁡A)=10a+3b0010a+3b-----1 And AAT=5a−b325a3−b2AAT=25a2+b215a−2b15a−2b13∵A(adj⁡A)=AAT∴10a+3b0010a+3b=25a2+b215a−2b15a−2b13[ using Eqs. (i) and (ii)]⇒15a−2b=0⇒a=2b15⋯⋯⋯...2 And 10a+3b=13.---3On substituting the value of ‘a’ from Eq.(iii) in Eq.(iv), we get 102b15+3b=13⇒20b+45b15=13⇒65b15=13⇒b=3Now , substituting the value of b in Eq. (iii) , we get 5a=2  Hence, 5a+b=2+3=5