Sum of Elements of Inverse of Matrix with Column of Ones

Theorem
Let $\mathbf B = \sqbrk b_n$ denote the inverse of a square matrix $\mathbf A$ of order $n$.

Let $\mathbf A$ be such that it has a row or column of all ones.

Then the sum of elements in $\mathbf B$ is one:


 * $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij}  = 1$

Proof
If ones appear in a row of $\mathbf A$, then replace $\mathbf A$ by $\mathbf A^T$ and $\mathbf B$ by $\mathbf B^T$.

Assume $\mathbf A$ has a column of ones.

Apply Sum of Elements of Invertible Matrix to the inverse $\mathbf B = \mathbf A^{-1}$:


 * $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$

where $\mathbf J_n$ denotes the square ones matrix of order $n$.

If $\mathbf A = \mathbf B^{-1}$ has a column of ones, then $\mathbf B^{-1} - \mathbf J_n$ has a column of zeros, implying determinant zero.

Substitute $\map \det {\mathbf B^{-1} - \mathbf J_n} = 0$ in Sum of Elements of Invertible Matrix:


 * $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - 0$

which implies the statement.