 Solution

Use banded Gaussian elimination followed by back substitution. To distinguish the right-hand side from the upper-diagonal coefficients $b_i$, write it as $r_i$. If the modified pivot elements are nonzero, the tridiagonal elimination recurrences areThey eliminate just the one entry below each pivot element. Solve the resulting upper bidiagonal matrix system byThere is a bounded number of arithmetic operations per row in each sweep, giving $O(n)$ arithmetic operations and $O(n)$ storage. The unpivoted recurrences cost $8n-7$ arithmetic operations under the usual scalar-operation count.

For a general tridiagonal matrix, nonsingularity alone does not guarantee these pivot elements. If a zero or unsafe pivot element occurs, use partial pivoting between the current row and the next row. These are the only possible nonzero candidates in the current column; adjacent row swaps introduce at most a second upper diagonal. Tracking that fill-in keeps both Gaussian elimination and back substitution at $O(n)$ arithmetic operations. For example, $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible but requires a row swap. If the input is singular, report a zero final pivot element; a unique solution cannot be promised, and compatibility or free-variable analysis is a separate issue. No nonsingularity or safe-pivot assumption is silently inferred from the word tridiagonal.