Question

let A and P be square matrices, with P invertible. Show that det(PAP^-1)= DetA

Answer 1

Well P*P^-1 is the n x n identity matrix I. IA is still A. Anything multiplied with the identity matrix remains the same. so det A = det A

Answer 2

You should know that
\[\det(AB) = \det(A) . \det(B)\]
and
\[\det(A^{-1}) = 1/\det(A)\]

Knowing those two things, it should be straightforward: take the left hand side and apply the first rule I just wrote down, then the second.

Answer 3

@cows: no, that assumes the matrix multiplication is commutative, which is false.

Answer 4

so its Det(P)*Det(P^-1)Det(A)=Det(A)
Det(I)Det(a)=det(a)

Answer 5

How do you know det(A^-1) is the same as 1/det(a)? What if det(a) = 0?

Answer 6

\[\det(PAP^{-1}) = \det(P) . \det(A) . \det(P^{-1}) = \det(P) . \det(A) . 1/\det(P) = \det(A)\]

Answer 7

@cows: if a matrix is invertible, it necessarily has non-zero determinant.