Starting with a diagonally dominant tridiagonal $n \times n$ matrix A = \begin{bmatrix} a & b & 0 & 0 & ... & 0 \\ b & a & b & 0 &... & 0 \\ 0 & b & a & b & 0 & 0 \\ 0 & 0 & \ddots & \ddots & \ddots & 0\\ 0 & 0 & 0 & b & a & b\\ 0 & 0 & 0 & 0 & b & \frac{a}{2} \end{bmatrix}
I want to show the linear system Ax=b, where b = \begin{bmatrix} c\\c\\c\\c\\c\\\frac{c}{2} \end{bmatrix} has a unique solution. I know how to find the determinant for a symmetric tridiagonal matrix but I don't the steps for when the main diagonal values are not the same. I can't think of a simple way to show the column vectors are linearly independent and since A isn't strictly diagonally dominant, I can't use any of the conclusions.
Let $T_0 = 1$ and $T_1 = a/2$ and for all other $n$, let $T_n$ denote the determinant of the $n \times n$ matrix $A$. Expanding along the first row, it is easy to see that the determinant satisfies the following recursion relation: \begin{align} T_n &= a T_{n-1} - b^2 T_{n-2} \\ T_0 &= 1 \\ T_1 &= a/2. \end{align} The solution is $T_n = 2^{-(n+1)} \left[ \left(a + \sqrt{a^2 - 4b^2} \right)^n + \left(a - \sqrt{a^2 - 4b^2} \right)^n \right]$. From what I gather from the question and comment, $A$ is diagonally dominant apart from the last row, which implies that $|a| \geq |2b|$. I'll also boldly assume that $a>0$ (else, repeat the analysis for $-A$). It follows that $T_n > 0$ for all $n$, and thus $A$ is invertible. Uniqueness of the solution follows.