Sum of Elements of Inverse of Matrix with Column of Ones

Theorem
Let $B =\paren { b_{ij} }$ denote the inverse of some $n\times n$ matrix $A$.

Assume $A$ has a row or column of all ones.

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


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

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

Assume $A$ has a column of ones.

Apply Sum of Elements of Invertible Matrix to invertible matrix $B = A^{-1}$:


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

Symbol $J_n$ from Sum of Elements of Invertible Matrix is the matrix of all ones.

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

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


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

which implies the statement.