TITLE: How to calculate the determinant of this n by n matrix?
QUESTION [5 upvotes]: Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots  & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new to matrices and determinants, but this is what I did:
I developed the determinant using the second column:
$$
(-1)^2*x_1 
\begin{pmatrix}
x_1  & 0 & \cdots & 0 \\
x_2  & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k  & 0 & \cdots& 1 \\
\end{pmatrix} + (-1)^3 *1
\begin{pmatrix}
0 & x_2 & \cdots & x_k \\
x_2  & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k  & 0 & \cdots& 1 \\\end{pmatrix}
$$
the first determinant is triangular, so its equal to $ x_1 $ but this is where I got stuck. I don't know what to do with the second determinant. Any help is appriciated. Thanks

REPLY [2 votes]: You should just factorize:\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots  & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix} = \begin{equation}\begin{pmatrix}
-1 & x_1& \cdots &x_k\\
0 &1 &\cdots &0\\
\vdots &&1\\
0&&&1
\end{pmatrix}\begin{pmatrix}
\sum x_i^2 & 0 &\cdots&\\
x_1&1&&\\
\vdots&&1&&\\
x_k&&&1
\end{pmatrix}
\end{equation}
and then \begin{equation}
\det\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots  & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}=\det\begin{pmatrix}
-1 & x_1& \cdots &x_k\\
0 &1 &\cdots &0\\
\vdots &&1\\
0&&&1
\end{pmatrix}\det\begin{pmatrix}
\sum x_i^2 & 0 &\cdots&\\
x_1&1&&\\
\vdots&&1&&\\
x_k&&&1
\end{pmatrix}
\end{equation}
\begin{equation}
=-1\cdot\sum x_i^2=-\sum x_i^2
\end{equation}