Calculate the determinant of the following n x n symmetric matrix:
Hint. Consider the second-order differences of the rows. When $n\ge3$, $$ \pmatrix{1&-2&1\\ &1&-2&1\\ &&\ddots&\ddots&\ddots\\ &&&1&-2&1\\ &&&&1&0\\ &&&&&1}A_n =\pmatrix{0&2\\ \vdots&0&\ddots\\ \vdots&\ddots&\ddots&\ddots\\ 0&\cdots&\cdots&0&2\\ n-2&n-3&\cdots&1&0&1\\ n-1&n-2&\cdots&\cdots&1&0}. $$
You can take this approach:
$$A_n=\begin{bmatrix}
0&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
2&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix} $$
Now, add the last column to the first one, notice it will always be equal to $n-1$. $(C_1 = C_1+C_n)$
$$\begin{bmatrix}
n-1&1&2& .&.&. &n-1\\
n-1&0&1&2& .&.&n-2 \\
n-1&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix} =(n-1)\begin{bmatrix}
1&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
1&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
1&n-2&.&.&2&1&0
\end{bmatrix}$$
From here, we can do as follows:
go from the last row towards the first and decrease each row's value with the one above it (for any row but the first one).
($\forall i \neq 1, R_i=R_i-R_{i-1}$, Starting with $i=n$ then $i=n-1 ... i=2$)
$$ = (n-1)\begin{bmatrix}
1&1&2& .&.&. &n-1\\
0&-1&-1&-1& .&.&-1 \\
0&1&-1&-1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&-1 \\
.&.&.&.&.&.&-1 \\
0&1&.&.&1&1&-1
\end{bmatrix}$$
Expand $C_1$:
$$ = (n-1)\begin{bmatrix}
-1&-1&-1& .&.&-1 \\
1&-1&-1&.&.&. \\
.&.&.&.&.&. \\
.&.&.&.&.&-1 \\
.&.&.&.&.&-1 \\
1&.&.&1&1&-1
\end{bmatrix}$$
Now Add the first row to all of the other rows ($\forall i \neq 1, R_i = R_i + R_1$)
$$ = (n-1)\begin{bmatrix}
-1&-1&-1& .&.&-1 \\
0&-2&-2&.&.&. \\
.&.&-2&.&.&. \\
.&.&.&.&.&-2 \\
.&.&.&.&.&-2 \\
0&.&.&0&0&-2
\end{bmatrix} = (n-1)[-1*(-2)^{n-2}]$$