An interval matrix family can contain a singular member

For each $t\in [-1,1]$, $\det A(t)=1-t^2\in [0,1]$. The direct interval product $[-1,1]\cdot [-1,1]=[-1,1]$ encloses $t^2$, so the Leibniz interval expression $1-[-1,1]=[0,2]$ contains every determinant. At $t=1$, $A(1)(1,-1{)}^{\mathsf{T}}=0$ with a nonzero vector, so $A(1)$ is singular. The interval $[0,2]$ includes zero and is not $\{0\}$; [F1] therefore makes no uniform nonsingularity claim from this enclosure.

If uniform residual data with $q<1$ existed for this whole family and any nonempty right-hand-side set, [F2] would make $A(1)$ invertible. This contradicts the nonzero null vector just exhibited. Hence such a usable uniform residual certificate cannot hold for the family.