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The midpoint quadratic spline collocation matrix isIts strict diagonal dominance margin is at least $3/4-2/8=1/2$. The inverse infinity-norm bound from diagonal dominance thus already proves $\Vert A_x^{-1}{\Vert }_{\infty }\le 2$: choosing an index where a vector $c$ attains its maximum absolute entry gives $\Vert A_{x}c{\Vert }_{\infty }\ge \frac{1}{2}\Vert c{\Vert }_{\infty }$, which also proves invertibility.

To evaluate the finite-dimensional norm exactly, set $D_{ii}=(-1{)}^i$ and $B=D{A}_{x}D$. If $T$ is the matrix with ones on its two adjacent diagonals and zeros elsewhere, then $B=(3/4)(I-T/6)$. Since $\Vert T/6{\Vert }_{\infty }\le 1/3$, the Neumann serieshas nonnegative entries. Thus $A_x^{-1}=D{B}^{-1}D$ has absolute entry values equal to those of $B^{-1}$. Its absolute row-sum vector is $v=B^{-1}1$, characterized byThe homogeneous roots are $3\pm 2\sqrt{2}$, and the constant particular solution is $v_i=2$. Writing $q=3-2\sqrt{2}\in (0,1)$ and imposing the two boundary values givesThis is largest at the middle row or rows. Therefore, with $i_{\ast }=\lfloor (n+1)/2\rfloor$, the exact norm isFor example, it is $4/3$ for $n=1$ and $8/5$ for $n=2$, and it increases towards two as the system grows. Thus two is a uniform bound, rather than the exact finite-$n$ inverse norm.