determinant of symmetric matrix of size (n+1)(n+1)

romsek

Math Team
Sep 2015
2,762
1,541
USA
$Det(a_0, d, n) =(-1)^{n - 1} (2^{n - 1} a_0 d^{n - 1} + (n - 1) 2^{n - 2} d^n)$

Now see if you can prove this via induction
 
  • Like
Reactions: 1 person
Jun 2016
25
2
Hong Kong
$\begin{vmatrix}
a_0 & a_1 & a_2 & \ldots & a_n\\
a_1 & a_0 & a_1 & \ldots & a_{n-1}\\
a_2 & a_1 & a_0 & \ldots & a_{n-2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_n & a_{n-1} & a_{n-2} & \ldots & a_0
\end{vmatrix}
=\begin{vmatrix}
-d & -d & -d & \ldots & a_n\\
d & -d & -d & \ldots & a_{n-1}\\
d & d & -d & \ldots & a_{n-2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
d & d & d & \ldots & a_0
\end{vmatrix}$
$=d^n\begin{vmatrix}
-1 & -1 & -1 & \ldots & a_n\\
1 & -1 & -1 & \ldots & a_{n-1}\\
1 & 1 & -1 & \ldots & a_{n-2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \ldots & a_0
\end{vmatrix}
=d^n\begin{vmatrix}
-1 & 0 & 0 & \ldots & a_n\\
1 & -2 & 0 & \ldots & a_{n-1}\\
1 & 0 & -2 & \ldots & a_{n-2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 0 & 0 & \ldots & a_0
\end{vmatrix}$
$=(-2)^{n-1} d^n\begin{vmatrix}
-1 & 0 & 0 & \ldots & a_n\\
1 & 1 & 0 & \ldots & a_{n-1}\\
1 & 0 & 1 & \ldots & a_{n-2}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 0 & 0 & \ldots & a_0
\end{vmatrix}
=(-2)^{n-1} d^n\begin{vmatrix}
-1 & a_n\\1 & a_0\end{vmatrix}
=2^{n-1} (-d)^n (2a_0+dn)$