Find the determinant of $I + A$ What is the determinant of an $n \times n$ matrix $B=A+I$, where $I$ is the $n\times n$ identity matrix and $A$ is the $n\times n$ matrix 
$$
A=\begin{pmatrix}
a_1 & a_2 & \dots & a_n \\
a_1 & a_2 & \dots & a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_1 & a_2 & \dots & a_n \\
\end{pmatrix}
$$
 A: Here is a proof using eigenvalues, wishing you have already seen this notion.
Clearly, as all the columns of $A$ are proportional, the kernel has dimension $n-1$, thus eigenvalue $0$ is of order $n-1$.
The sum of the eigenvalues being equal to the trace $\sum_i a_i$, the $n$-th eigenvalue is:
$$\lambda:=\sum_i a_i.$$
Thus as the spectrum of $A+xI$ is equal to the spectrum of $A$ shifted by x (here with $x=1$), it is constituted of $1$ (with multiplicity $n-1$) and $\lambda+1$.
The determinant of a matrix being equal to the product of its eigenvalues, the final result is:
$$\det(A+I)=1+\sum_i a_i$$
A: Using the Weinstein-Aronszajn determinant identity,
$$\det \left({\bf I}_n + {\bf 1}_n {\bf a}^{\top} \right) = \det \left( 1 + {\bf a}^{\top} {\bf 1}_n \right) = 1 + {\bf a}^{\top} {\bf 1}_n = 1 + \sum_{i=1}^n a_i$$
A: First look at what happens when $n=2$, $n=3$, renaming our matrix $B_n(a_1,\cdots,a_n)$ one gets
$$\begin{align}\det{B_2(a_1,a_2)}&=1+a_1+a_2\\\det{B_3(a_1,a_2,a_3)}&=1+a_1+a_2+a_3\end{align}$$
Assume that $\det{B_{n-1}(a_1,\cdots,a_{n-1})}=1+a_1+\cdots a_{n-1}$ and consider
$$\begin{vmatrix}1+a_1    &  a_2   & \cdots & a_n\\
                   a_1    & 1+a_2  & \cdots & a_n\\
                   \vdots & \vdots & \ddots & \vdots\\
                   a_1    &   a_2  & \cdots & 1+a_n
\end{vmatrix}$$
Substract the second row from the first to get
$$\begin{vmatrix}  1      &  -1    & 0 & \cdots & 0\\
                   a_1    & 1+a_2  & a_3 &\cdots & a_n\\
                   a_1 & a_2 & 1+a_3 & \cdots & a_n\\
                   \vdots & \vdots & \vdots & \ddots & \vdots\\
                   a_1    &   a_2  & a_3 &\cdots & 1+a_n
\end{vmatrix}$$
Then add the first column to the second and get
$$\begin{vmatrix}  1      &  0    & 0 & \cdots & 0\\
                   a_1    & 1+a_1+a_2  & a_3 &\cdots & a_n\\
                   a_1 & a_1+a_2 & 1+a_3 & \cdots & a_n\\
                   \vdots & \vdots & \vdots & \ddots & \vdots\\
                   a_1    &   a_1+a_2  & a_3 &\cdots & 1+a_n
\end{vmatrix}$$
Developing along the first row one gets
$$\det{B_n(a_1,\cdots,a_n)}=\det{B_{n-1}(a_1+a_2,\cdots ,a_n)}$$
And this by the induction assumption is $$1+a_1+a_2+\cdots+a_n$$
