# Sum of Elements of Invertible Matrix

## Theorem

Let $\mathbf J_n$ be the $n \times n$ square ones matrix.

Let $\mathbf B$ be an $n\times n$ invertible matrix with entries $b_{i j}$, $1 \le i, j \le n$.

Then:

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

## Proof

### Lemma

Let$\mathbf J_n$ be the $n \times n$ square ones matrix.

Let $\mathbf A$ be an $n \times n$ matrix.

Let $\mathbf A_{ij}$ denote the cofactor of element $a_{ij}$ in $\map \det {\mathbf A}$, $1 \le i, j \le n$.

Then:

 $\ds \map \det {\mathbf A - \mathbf J_n}$ $=$ $\ds \map \det {\mathbf A} - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{i j}$ Cofactor Sum Identity

$\Box$

Let $\mathbf A = \mathbf B^{-1}$.

Then:

 $\ds \mathbf A \mathbf B$ $=$ $\ds \mathbf I$ Definition of Invertible Matrix $\ds \mathbf B$ $=$ $\ds \dfrac {\adj {\mathbf A} } {\map \det {\mathbf A} }$ Matrix Product with Adjugate Matrix with $\mathbf B = \mathbf A^{-1}$

Then:

 $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij}$ $=$ $\ds \dfrac 1 {\map \det {\mathbf A} } \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij}$ $\ds$ $=$ $\ds \map \det {\mathbf B} \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij}$ Determinant of Matrix Product and $A B = I$ $\ds$ $=$ $\ds \map \det {\mathbf B} \paren {\map \det {\mathbf A} - \map \det {\mathbf A - \mathbf J_n} }$ Apply the Lemma $\ds$ $=$ $\ds \map \det {\mathbf B} \map \det {\mathbf A} - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$ Definition $A = B^{-1}$ $\ds$ $=$ $\ds 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$ Determinant of Matrix Product and $\mathbf B \mathbf A = \mathbf I$

$\blacksquare$