Determinant of rank-$1$ update of multiple of identity matrix I've got to calculate determinant for such matrix:
$$ \begin{bmatrix}
a_1+b & a_2 & \cdots & a_n\\
a_1 & a_2+b & \cdots & a_n\\
\vdots & \vdots & \ddots & \vdots\\ 
a_1 & a_2 & \cdots & a_n+b\\
\end{bmatrix} $$
Please give me some tips how to calculate this.
 A: Set $A=\sum\limits_{i=1}^na_i$. By multilinearity,
\begin{align*}
&\begin{vmatrix}
a_1+b &a_2&\dots&a_n\\
a_1&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
a_1  a_2&\dots &a_n+b
\end{vmatrix}=
\begin{vmatrix}
 A+b &a_2&\dots&a_n\\
A+b&b+a_2  &\dots&a_n \\
\vdots&\vdots&&\vdots\\
A+b & a_2&\dots &b+a_n 
\end{vmatrix}\\[1ex]
&=(A+b)\begin{vmatrix}
 1 &a_2&\dots&a_n\\
1&b+a_2  &\dots&a_n \\
\vdots&\vdots&&\vdots\\
1 & a_2&\dots &b+a_n 
\end{vmatrix}=(A+b)\begin{vmatrix}
 1 &a_2&\dots&a_n\\
0&b &\dots& 0 \\
\vdots&\vdots&&\vdots\\
0 & 0&\dots &b
\end{vmatrix}\\[1ex]
&=\color{red}{(A+b)b^{n-1}}.
\end{align*}
A: Write the matrix as $A+bI.$  Here, all the rows of $A$ are the same, and so $A$ is rank $1$, and therefore the kernel is of dimension $n-1$ and there is only one non-trivial eigenvalue, $\operatorname{tr}(A)$.  Therefore the characteristic polynomial of $A$ is $p_t(A)=\det(tI-A)=t^{n-1}(t-\operatorname{tr}(A))$.
It is now straightforward to calculate $\det(A+bI)$ from $\det(tI-A)$.  
A: The structure of this matrix aloud to write this equation which does not generally hold
$$det(A+Ib)=b^{n-1}(tr(A)+b)$$

I guess that's related with the fact that
$$A^2=tr(A)A$$
but just now I don't see how... If someone can see just edit in the comments please
A: The matrix can be written as $A+bI$, where $A$ is the matrix with all rows equal to $a_1,a_2,\dots,a_n$. The determinant in question is $(-1)^n\chi_A(-b)$, where $\chi_A$ is the characteristic polynomial of $A$.
Since $A$ has rank $1$, we have $\chi_A(x)=x^n-tr(A)x^{n-1}$ (see this for instance).
Finally, the determinant in question is $(-1)^n\chi_A(-b)=(-1)^n(-b)^n-a(-b)^{n-1}=b^n+ab^{n-1}$.
