Let A be a matrix of order $n$, where the sum of each column's elements adds to $n$, and $v_n = (1,...,1)$. Prove $\ |A|=\sum_{i=1}^n |A_i|$

Question

I have the following question for my assignment, I am only asking for direction as I am stuck:

Given a square matrix $A$ of order $n$, where the sum of each column's elements adds to $n$, and row vector $v_n = (1,...,1)$. Prove or disprove: $$\ |A|=\sum_{i=1}^n |A_i|,$$ where $A_i$ is the matrix obtained from $A$ by replacing its $i$th row with the vector $v_n$.

I have tried to use that $|A + B| = |A| + |B|$ if $A$ and $B$ differ exactly by one row, but I am getting stuck. I also thought of using Cramer's rule, but am struggling with how to use that.

Answer 1

Note, every $|A_i|$ can be obtained by adding all the rows to the $i$th row and then dividing that row by $n$. We got $|A_1|+\dots+|A_n|=1/n|A|+\dots+1/n|A|=|A|$.

Answer 2

Yes, you may use Cramer’s rule when $n$ is a unit in the underlying field.

Let $x$ be the vector of ones and $b=nx$. By assumption, we have $x^TA=b^T$. Therefore, by Cramer’s rule, we have $x_i|A|=|B_i|$ for each $i$, where $B_i$ is the matrix obtained by replacing the $i$-th row of $A$ by $b^T$. Hence $|B_i|=n|A_i|$. In turn, we obtain $|A|=x_i|A|=|B_i|=n|A_i|$. Therefore $n|A|=n\sum_i|A_i|$ and in a field where $n$ is a unit, we have $|A|=\sum_i|A_i|$.