Let A and B be 3 times 3 matrices such that det (A) = 2 and det (B) Compute: det(3A), det(AB), det(A^-1). Solution det(A) = 2 det(B) = 3 det(3A) Multiplying A by 3 is the same as multiplying each row of A by 3. Since A has 3 rows and multiplying a row by a constant multiplies the determinant by the same constant, det 3A = 33 det A = 33 (2) = 9(2)=18 Therefore det(3A)=18 det(AB)= det(A)det(B) = 2*3=6 therefore det(AB) = 6 det(A-1)= det(A) therefore det(A-1) = 2.