Question

Suppose that A = PDP-1 . Prove that det(A) = det(D)

Answer 1

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that  for two matrices A and B we have that:

det(AB) = det(A)*det(B) and $det(A^{-1}) = \frac{1}{det(A)}$. Now:

$A = PDP^{-1}$

$det(A) = det(PDP^{-1})$

$det(A) = det(P)*det(D)*det(P^{-1})$

$det(A) = det(P)*det(D)*\frac{1}{det(P)}$

$det(A) = det(P)*\frac{1}{det(P)}*det(D)$

$det(A) = det(D)$.