# Inverse of diagonally dominant matrix with equal off-diagonal entries

Is there an explicit expression for the inverse of strictly diagonally dominant matrix with identical off-diagonal elements?

For example:

$$\begin{pmatrix} a & -b & -b \\ -b & c & -b \\ -b & -b & d \end{pmatrix}$$

where $|a|\gt 2|b|$, $|c|\gt 2|b|$, and $|d|\gt 2|b|$. I'm looking for an inverse for an arbitrary $n\times n$ matrix with the above property.

• This can be expressed as the inverse of a rank one update of a diagonal matrix, a topic that has been addressed in previous questions. – hardmath Feb 3 '15 at 23:01
• Any recommended references that can help? – user212801 Feb 3 '15 at 23:45
• – Martin Sleziak Oct 31 '16 at 13:12

The Sherman-Morrison formula gives the inverse of any rank one update of a matrix for which the inverse is known.

Here we can write your matrix as follows:

$$\begin{pmatrix} a & -b & -b \\ -b & c & -b \\ -b & -b & d \end{pmatrix} = \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix} - b \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$$

Since the first summand is an invertible diagonal matrix (regardless of the actual sign of $b$), we have that the Sherman-Morrison formula can be applied.

Let $A = \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix}$ and let $u = \begin{pmatrix} -b \\ -b \\ -b \end{pmatrix}$, $v^T = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix}$. What you ask for is:

$$(A + uv^T)^{-1} = A^{-1} - \left( \frac{1}{1+ v^T A^{-1} u} \right) \left( A^{-1} uv^T A^{-1} \right)$$

Note that the first factor in the second term of the right hand side is just a scalar, obtained by taking the reciprocal of the scalar $1+ v^T A^{-1} u$. It multiplies the rank one matrix $A^{-1} uv^T A^{-1}$, showing that the inverse of the rank one update of $A$ is itself a rank one update of $A^{-1}$.