|Oct22-07, 04:24 PM||#1|
[b]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?
|Oct22-07, 04:53 PM||#2|
We know that det(AB) = det(A)det(B).
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.
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