i Let A be a 3 times 3 matrix with determinant 15 Then detad

(i) Let A be a 3 times 3 matrix with determinant -15. Then det(adj(A^T)) = -15, det(adj(A^-1) = -1/15 and det(adj(7A) = (ii) Let A be a 2 times 2 matrix such that adj(A) = [1 5 2 -7] Then det A =

Solution

1.

Det(A) = -15

det(adj(A^T)) = -15

we know that

det(adj(A)) = det(A)^(n-1)

det(kA) = k^n*det A

here n = 3

det(adj A) = det(A)^2 = -15^2 = 225

det (adj(7A)) = 7^3*det(adj(A)) = 7^3*225 = 77175

B.

det(adj(A)) = det(A)^(n-1)

here n = 2

det(adj(A)) = det(A)^(2-1)

det(adj(A)) = det(A)

det(A) = -7*1 - 5*2 = -17