Question

Let and P be square matrices with invertible. Show that det (PAP-1) det A

Rewrite det (PAP - 1) as an expression containing det Choose the correct answer below
a. det (PAP-1) = [(det PYdet A)(det P-1)]-1
b. det t(PAP -1)= det P+ det A det P-1
c. det (PAP -1) = (det PI(det AJ(det P - )
d. det (PAP-1) = (det P +det A+det P -1) -
The expression det P - can be rewritten in terms of det P Rewrite det P - in terms of det P
det P - (det P)
Using the Commutative Property and the equation in the previous step; the expression from the first step can be rewritten as ((det PI(det P) (det A)
This new expression simplifies to det because (det P)det P) Thus det (PAP - 1 ) = det A

Answer 1

Answer: c. det(PAP^-1) = det(A)

To show that det(PAP-1) = det(P)det(A)det(P^-1), we can use the following properties of determinants:

1. det(AB) = det(A)det(B) for any matrices A and B

2. det(A^-1) = 1/det(A) for any invertible matrix A

Starting with det(PAP^-1), we can apply property 1 to get:

det(PAP^-1) = det(P)det(A)det(P^-1)

Then, we can use property 2 to rewrite det(P^-1) as:

det(PAP^-1) = det(P)det(A)(1/det(P))

Simplifying this expression gives:

det(PAP^-1) = det(A)

Therefore, det(PAP^-1) = det(P)det(A)det(P^-1) is equivalent to det(PAP^-1) = det(A).

So, the correct answer is:

c. det(PAP^-1) = det(A)