Understanding Determinants: Proof of det(kA) = k^ndetA and det(rI(n)) = r^(n)

[captainjack2000]

1. The proof that det(kA) = k^ndetA where A is nxn
I read somewhere that det(rI(n)) = r^(n)
so det(rA) = det(rI(n).A) = r^ndetA but I am really confused about how they got that? Is I the identity matrix? What would the det(I) be?

 

[JasonRox]

1. The proof that det(kA) = k^ndetA where A is nxn
I read somewhere that det(rI(n)) = r^(n)
so det(rA) = det(rI(n).A) = r^ndetA but I am really confused about how they got that? Is I the identity matrix? What would the det(I) be?

You pretty much got it.

We know that det(AB) = det(A)det(B).

Therefore,

det(rA) = det(rI[n]A) = det(rI[n])det(A) = r^n det(I[n]) det(A) = r^n (1) det(A) = r^n det(A)

I showed every step possible basically.

Yes, the determinant of I[n] is always 1. It is the identity matrix.