The actual problem:
In each row of an $n \times n$ matrix, the sum of elements equals $2008$. Prove that if we replace all values of one row by $1$-s, the determinant becomes $\frac{1}{2008}$ times the original determinant.
How could I prove this?
What I tried is applying Laplace expansion on both the original and the modified matrix. So I got something like
\begin{align} \det(A) &= a_{1,1} [\ldots] - a_{1,2} [\ldots] + \ldots + a_{1,n} [\ldots] \tag{1} \\ \det(A') &= [\ldots] - [\ldots] + \ldots + [\ldots] \tag{2} \end{align}
Where $A$ is the original matrix and $A'$ is the modified matrix. I replaced the first row by $1$-s for convenience and expanded by the first row. The $[\ldots]$-s are the minors, which match in the two equations. What we know is that
$$ a_{1,1} + a_{1,2} + \ldots + a_{1,n} = 2008 $$
Because $a_{1,j}$ are all the elements of the first row.
What we essentially have to prove is that $\frac{\det(A)}{\det(A')} = 2008 \iff \frac{\det(A')}{\det(A)} = \frac{1}{2008}$
How do I continue? Should I do something like work with $(1) \div (2)$? Is this approach formal enough? What other approaches would be possible?
I believe this claim is incorrect, for example: $\left[\begin{matrix} 1 \ 2007 \\ 1 \ 2007 \end{matrix}\right]$ is singular, but obviously if you replace any row with all 1's it becomes regular.
However, there is a similar theorem that does hold: if the sum in each column is 2008, then you can replace any row with 1's and the determinant of the resulting matrix will be $\frac {1}{2008} $ of the determinant of the original matrix. Proof: take the original matrix, and then: