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Consider the original small Poisson equation exampleIts exact solution is the quadratic polynomial $u(x)=1+2x-x^2$. Use a uniform mesh $x_i=ih$ with $h=1/4$ and three interior unknowns $U_1,U_2,U_3$. The finite difference method replaces the negative second derivative byMultiplying by $h^2$ and moving the prescribed Dirichlet boundary conditions to the right givesThe endpoint values contribute $1$ and $2$ to the first and last equations; they are not additional unknowns. The Thomas algorithm givesand back substitution yieldsAll three values equal $u(x_i)$ exactly: the centered finite difference for the second derivative is exact on a quadratic polynomial. This special nodal exactness does not assert exactness for arbitrary forcing or for the piecewise-linear curve joining the nodes.

For any $n$ interior points with $h=1/(n+1)$, the scaled tridiagonal matrix has $b_i=2$ and $a_i=c_i=-1$. Its pivot elements are $p_i=(i+1)/i$: start at $p_1=2$ and use $p_i=2-1/p_{i-1}$. All are positive. Interior rows have equality in diagonal dominance, so strict diagonal dominance is not the reason this example works. Instead it is a symmetric positive-definite matrix, since for a real vector $v\ne 0$ with $v_0=v_{n+1}=0$,Zero would force every neighbouring value equal and hence all values zero. This connects the boundary constraints directly to the safe pivots for tridiagonal elimination.