= Solution
For order six, the support-midpoint sites are $x_i=i+3$, so
$$
(A_x)_{ij}=N_{j,6}(i+3)=N_{0,6}(i+3-j),\qquad0\le i,j\le n.
$$
By part (b), the complete finite <matrix> is the symmetric five-diagonal <matrix>
$$
\boxed{(A_x)_{ij}=\frac1{120}\begin{cases}
66,&i=j,\\26,&|i-j|=1,\\1,&|i-j|=2,\\0,&|i-j|>2.
\end{cases}}
$$
Only indices from $0$ through $n$ are retained; the first and last rows are truncated, with no exterior <coefficients> or wraparound terms. For a row with all possible neighbors, the <strict diagonal dominance> margin is $(66-26-26-1-1)/120=1/10$. Boundary rows have an equal or larger margin.
Here is the appropriate inverse estimate with its proof. If a square <matrix> $A$ has $|a_{ii}|-\sum_{j\ne i}|a_{ij}|\ge d>0$, choose an index $i$ maximizing $|z_i|$. The <reverse triangle inequality> gives
$$
|(Az)_i|\ge|a_{ii}|\,|z_i|-\sum_{j\ne i}|a_{ij}|\,|z_j|\ge d\|z\|_{\ell^\infty}.
$$
Thus $Az=0$ implies $z=0$, and a square finite <matrix> is invertible. Applying the same estimate with $z=A^{-1}b$ proves $\|A^{-1}b\|_{\ell^\infty}\le d^{-1}\|b\|_{\ell^\infty}$. The <inverse infinity-norm bound from diagonal dominance> consequently yields
$$
\boxed{\|A_x^{-1}\|_{\ell^\infty}\le10.}
$$
This also proves existence and uniqueness of the requested interpolating spline for every data vector and every finite $n$. The <quintic cardinal spline midpoint collocation> <matrix> includes the cases of one or two <coefficients>, where the dominance margin is even larger.
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