Inverse matrix of matrix (all rows equal) plus identity matrix Let $A$ be a matrix where all rows are equal, for example,
$$A=\left[\begin{array}{ccc}
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3}
\end{array}\right]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=\left[\begin{array}{ccc}
a_{1}+1 & a_{2} & a_{3} \\
a_{1} & a_{2}+1 & a_{3} \\
a_{1} & a_{2} & a_{3}+1
\end{array}\right]$$
I have a conjecture, which computation has so far confirmed:
$$B^{-1}=I-\frac{A}{\mbox{tr}(A)+1}$$
Why is this true?
 A: In the following, let us assume that $\mbox{tr}(A)+1\not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)\left(I-\frac{A}{\mbox{tr}(A)+1}\right)$ and see if we get the identity matrix.  
$$C=-\frac{1}{\mbox{tr}(A)+1}A^2+\frac{\mbox{tr}(A)}{\mbox{tr}(A)+1}A+I$$  
so that to get the desired result, we need to prove that $A^2=\mbox{tr}(A)A$.  
The $i,j$ coefficient of $A^2$ is given by $$\sum_{k=1}^n A_{i,k}A_{k,j}=\sum_{k=1}^n a_ka_j=\mbox{tr}(A)a_j=\mbox{tr}(A)A_{i,j} $$  
So that indeed, we obtain the desired identity.  
NB : To justify that the matrix $B=A+I$ is invertible if and only if $\mbox{tr}(A)+1\not = 0$, note that the above computation actually establishes the identity $$(\mbox{tr}(A)+1)I=(A+I)((\mbox{tr}(A)+1)I-A)$$
from which the equivalence follows.
A: The inverse of $B$:
$$B^{-1}=\frac{\text{adj}{(B)}}{\det(B)}.$$
The determinant of $B$:
$$\det(B)=\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
a_{1} & a_{2}+1 & a_{3} \\
a_{1} & a_{2} & a_{3}+1\end{vmatrix}=
\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
-1 & 1 & 0 \\
-1 & 0 & 1\end{vmatrix}=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$\text{adj}(B)=\text{C}^T=\\
\begin{pmatrix}
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \\
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \\
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
\end{pmatrix}=\\
\begin{pmatrix}
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \\
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \\
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
\end{pmatrix}=\\
\begin{pmatrix}
a_1+a_2+a_3+1 & 0 & 0 \\
0 & a_1+a_2+a_3+1 & 0 \\
0 & 0 & a_1+a_2+a_3+1
\end{pmatrix}-
\begin{pmatrix}
a_1 & a_2 & a_3 \\
a_1 & a_2 & a_3 \\
a_1 & a_2 & a_3
\end{pmatrix}.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
A: $A=ea^T$ where $a=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ and $e = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if  $1+v^TC^{-1}u \ne 0$ and $$(C+uv^T)^{-1}=C^{-1}-\frac{C^{-1}uv^TC^{-1}}{1+v^TC^{-1}u}.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+\sum_{i=1}^3a_i=1+trace(A) \ne 0.$ and
$$B^{-1}=I-\frac{ea^T}{1+trace(A)}=I-\frac{A}{1+trace(A)}$$
