This comes up pretty often, I think it is worth throwing in a bit more than you asked.
Suppose I have an $n$ by $n$ square matrix where all entries are equal to $1$:
$$ R_n \; \; = \; \;
\left( \begin{array}{ccccccc}
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1
\end{array}
\right).
$$
What are the eigenvalues and eigenvectors? ( Here $n=7,$ I will keep typing $n$ because the exact value really does not matter).
Well, we get an automatic eigenvalue $n,$ with eigenvector (I use columns always)
$$
\left( \begin{array}{c}
1 \\
1 \\
1 \\
1 \\
1 \\
1 \\
1
\end{array}
\right).
$$
We then get $n-1$ genuine eigenvalues $0,$ with linearly independent eigenvectors
$$
\left( \begin{array}{c}
1 \\
-1 \\
0 \\
0 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
-1 \\
0 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
-1 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
-1 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
0 \\
-1 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
0 \\
0 \\
-1
\end{array}
\right) \;
$$
The complete list of eigenvalues is $\left\{n,0,\ldots,0 \right\}.$
What would I get for the eigenvalues of $k R_n,$ for a real number $k?$ With exactly the same eigenvectors, I would now have eigenvalues $\left\{kn,0,\ldots,0 \right\}.$ In your example the value of $k$ is $-1.$ All that has changed is the eigenvalue of the eigenvector with all 1's, which is now sent to $kn$ times itself.
What are the eigenvalues of $ w I + k R_n,$ with real numbers $w,k?$ With exactly the same (basis of) eigenvectors, we just add, for each eigenvector, $w$ times each. So the eigenvalues are $\left\{w + kn,w,\ldots,w \right\}.$ Again, these are genuine eigenvalues (the Jacobi normal form is diagonal, which we also know from symmetry). The determinant of the matrix is then $$ ( w + k n ) \; w^{n-1} = w^n + k \; n \, w^{n-1}.$$
When I have a basis of eigenvectors, and I know the list of eigenvalues, the characteristic polynomial is just the product of $n$ linear terms, each one given by $\left(\mbox{eigenvalue} - \lambda \right)$ in the order you are using. Here, your answer is just
$$ ( w + k n - \lambda) \; ( w - \lambda)^{n-1}.$$
Now, for your example, indeed $n=3.$
How to get the 3's on the diagonal? In order to get the $-1'$s off the diagonal, we must take $k=-1,$ as I mentioned. So we must add 4 to bump $-1$ up to 3, so that $w=4.$ Meanwhile $w + k n = 1$ and $n-1=2.$ Your characteristic polynomial is
$$ \det (A - \lambda I) = (1-\lambda) \; (4 - \lambda)^2.$$
Arturo and Didier (and all enlightened people) would want me to point out how spectacularly incorrect this method is for a matrix such as
$$
\left( \begin{array}{rr}
3 & 1 \\
0 & 3
\end{array}
\right) ,
$$
where the characteristic polynomial is easy enough to find but there is no basis of eigenvectors.