Characteristic polynomials of matrices related How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.
$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\ 
\end{bmatrix}$
$B=\begin{bmatrix}
-1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\ 
\end{bmatrix}$
Where $A$ and $B$ are symmetric matrices of order $n+1$.

About first matrix. 
Lets take the bottom line. As we know :
$$\det{A} = \sum{(-1)^{i+j}\cdot a_{i,j}\cdot M_{i,j}},$$ so we got:
$$S_{n+1} = a_{n+1,n+1}\cdot (-1)^{2n+2}\cdot S_{n} + (-1)^{n+2}a_{n+1,1}S'_{n},$$ where $S'_{n}= (-1)^{1 + n} \lambda^{n-1}(-1)^{n-1}$, because of :
$S'_{n} = \begin{bmatrix}
1 & 1 & \cdots & 1 & 1\\
-\lambda & 0 & \cdots & 0 & 0 \\
0 & -\lambda & \cdots & 0 & 0\\
\vdots & \vdots & \cdots &\ddots & \vdots \\
0 & 0 & \cdots & -\lambda & 0 \\ 
\end{bmatrix}$
