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:

  • Add all rows to the given row. This does not change the determinant, but all elements in that row will become column-wise sums, i.e. 2008.
  • Divide that row by 2008. That makes all values in that row equal to 1, and divides the determinant by 2008.
  • $\begingroup$ Gosh, I misread the problem... And I was trying to prove something that isn't even true... Yes, the problem said what you mention (although with the sum is $2008$ in each row and we replace a column, but I suppose that's essentially the same). Lesson: Always test whether a statement holds before you start attempting to prove it... $\endgroup$ – bertalanp99 Nov 29 '17 at 20:14

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