I am looking at this proof:
Let A be an n x n matrix and c a scalar. Show that det(cA)=c^ndet A.
Suppose it true for n=k, now consider any (k+1)x(k+1) matrix A, and expand
about the first row:
det(cA) = sum_{r=1 to k+1} (-1)^(r-1) c A_{1,r} det (cB)_{1,r}
where (cB)_{1,r} is the matric obtained from cA by deleting the 1st row and
the r-th col. Then by assumption:
det(cA) = sum_{r=1 to k+1} (-1)^(r-1) c^{k+1} A_{1,r} det (B)_{1,r}
...........= c^{k+1} det(A)
So if its true for n=k its true for n=k+1. It is trivialy true for n=1.
Hence we conclude by mathematical induction that it is true for all
positive integers.
RonL
Here is a theoerem that you can factor a multiple out of each row or colomun.
Say you want to
[ 2 4]
[3 5]
You can factor a 2 to get,
[1 2]
[3 5]
Then then find determinant of that and multiply by the number you factored out.
But, the matrix c*A is a matrix were all entries are multipled by c.
Thus, you factor c from each row, there are n rows, thus it is:
c*c*...*c --> n times.
Meaning,
c^n