The symmetric, tridiagonal $n-$by$-n$ matrix with the elements $a_{ii+1} = a_{i+1i} $ and off-diagonals' absolute values equal to the diagonal (except for row 1 and row n) is invertible. The elements on the diagonal are nonzero.
How can I go about showing this? Row reducing and finding the determinant would be too long.
Consider $$ A_n:=\mathrm{tridiag}(1,1,1), $$ that is, a tridiagonal matrix with constant entries on the three diagonals. Clearly, these matrices satisfy the given conditions. The eigenvalues of $A_n$ are $$ 1+2\cos\left(\frac{k\pi}{n+1}\right), \quad k=1,\ldots,n. $$ Hence, for any $n=3m-1$, $m=1,2,\ldots$, $A_n$ is singular.