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

#### mathodman25

Let for j = 0,. . . n aj = a0 + jd, where a0, d are fixed real numbers.
Calculate the determinant of the matrix A of size (n + 1) Ã— (n + 1)

#### Attachments

• 13.9 KB Views: 1

#### romsek

Math Team
$Det(a_0, d, n) =(-1)^{n - 1} (2^{n - 1} a_0 d^{n - 1} + (n - 1) 2^{n - 2} d^n)$

Now see if you can prove this via induction

• 1 person

#### fungarwai

$\begin{vmatrix} a_0 & a_1 & a_2 & \ldots & a_n\\ a_1 & a_0 & a_1 & \ldots & a_{n-1}\\ a_2 & a_1 & a_0 & \ldots & a_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_n & a_{n-1} & a_{n-2} & \ldots & a_0 \end{vmatrix} =\begin{vmatrix} -d & -d & -d & \ldots & a_n\\ d & -d & -d & \ldots & a_{n-1}\\ d & d & -d & \ldots & a_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ d & d & d & \ldots & a_0 \end{vmatrix}$
$=d^n\begin{vmatrix} -1 & -1 & -1 & \ldots & a_n\\ 1 & -1 & -1 & \ldots & a_{n-1}\\ 1 & 1 & -1 & \ldots & a_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \ldots & a_0 \end{vmatrix} =d^n\begin{vmatrix} -1 & 0 & 0 & \ldots & a_n\\ 1 & -2 & 0 & \ldots & a_{n-1}\\ 1 & 0 & -2 & \ldots & a_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & \ldots & a_0 \end{vmatrix}$
$=(-2)^{n-1} d^n\begin{vmatrix} -1 & 0 & 0 & \ldots & a_n\\ 1 & 1 & 0 & \ldots & a_{n-1}\\ 1 & 0 & 1 & \ldots & a_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & \ldots & a_0 \end{vmatrix} =(-2)^{n-1} d^n\begin{vmatrix} -1 & a_n\\1 & a_0\end{vmatrix} =2^{n-1} (-d)^n (2a_0+dn)$