So just to start I am not aware if this can be generalized or if it is just a weird occurrence. But I was doing by problem set for linear algebra and I had to compute the Wronskian determinant of the following functions. $$ f_1(x)= e^x, \quad f_2(x)=xe^x, \quad f_3(x) = x^2e^x $$ so I got the matrix: $$ W= \begin{bmatrix} e^x& xe^x& x^2e^x \\ e^x& e^x+xe^x& 2xe^x+x^2e^x \\ e^x& 2e^x+xe^x& 2e^x+4xe^x+x^2e^x \\ \end{bmatrix} $$
and what I noticed is that if if take an $e^x$ out of W and call the new matrix W' we get: $$ W'= e^x\begin{bmatrix} 1& x& x^2 \\ 1& 1+x& 2x+x^2 \\ 1& 2+x& 2+4x+x^2 \\ \end{bmatrix} $$
Now I noticed that $det(W)=e^{3x}det(W')$ and I was wondering that could this be generalized to all functions such that say we have again a 3x3 matrix, say $M$ and we have a common factor that is the analytic function say $f(x)$ then if we take $f(x)$ out of $M$, like you can for a constant, to get a new matrix $M'$ then would the same situation appear such that $det(M)=[f(x)]^3 det(M')$ ? In my example $f(x) = e^x$, so can this be true for any analytic function ?
Yes, this property can be generalized even more so than you specify. You can take ANY size NxN matrix and factor out ANY function f(x). You don't need a "common factor" since you can always pull out f(x) from g(x) as follows: g(x)=f(x)*[ g(x)/f(x) ]
So for a NxN matrix, M, with entries Mij at the ith row and jth column
det(M)=[f(x)]^N*det(M′) where the entries of matrix, M' are M'ij=Mij/f(x)
This is because multiplying the row of a matrix by a constant,K, multiplies the determinant by K. So dividing out f(x) from all N rows would multiply the determinant by 1/(f(x))^N