Let A be an nxn matrix and c a scalar. Show that det(cA) = c^n * det(A)
One property of the determinant is that if you multiply an entire row (or entire column) by some c then the new determinant is c*original. This is shown inductively: A = [a b] A' = [K*a K*b] det A = ad-bc [c d] [c d] det A' = Kad-Kbc = K*det A we then use the fact that the determinant of a 3x3 is found using 2x2 minors, or more generally (N+1)x(N+1) using NxN. This property is seen to remain true by induction. c*A where a is nxn is multiplying n rows by c, so if we remove a constant multiplication of a row one at a time we find c^n as a constant multiplied by the determinant, ie: det (cA) = c^n*det(A)
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