Let A and B be 3 times 3 matrices such that det A 2 and det

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