Can these determinants be analytically continued?

Question

Let $A$ be the matrix:

$$A(n,k): \text{If } k \text{ divides } n \text{ then } \frac{k}{n^{\frac{1}{4} (3+2 i t)}} \text{ else } 0 $$

and let $B$ be the matrix:

$$B(n,k): \text{If } k \text{ divides } n \text{ then } \frac{\mu (k)}{n^{\frac{1}{4} (3+2 i t)}} \text{ else } 0 $$

where $\mu(n)$ is the Möbius function.

Then let $d$ be the determinant of the matrix product of matrix $A$ times the transposed matrix $B$, plus the identity matrix (as in linear algebra) :

$$d(t) = \left|A.B^{\mathsf{T}}+\text{IdentityMatrix}\right|$$

The plot of the real part of the determinant $d(t)$ of the symmetric $17$x$17$ matrix looks like this:

where the black lines are at the Riemann zeta zeros on the critical line.

(I don't have the computing power to compute larger matrices than this.)

Compare this to the partial sums of the Möbius function times some constant plotted in red on the critical line, superimposed with the imposed with the determinant d(t) plotted in blue:

where the partial sum of the Möbius function is:

$$2.3 \; \Re\left(\sum _{n=1}^{\text{17}} \frac{\mu (n)}{n^{\frac{1}{2}+i t}}\right)$$

The constant 2.3 is not any specific number it is just there to make the height of the plots match somewhat.

One could then say that the determinant is equal to the characteristic polynomial of $A.B^{\mathsf{T}}$ evaluated at $x=-1$.

Because the reciprocal of Riemann zeta: $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$ is the analytical continuation of the Möbius function sum, can the determinant $d(t)$ be analytically continued to the critical line?

Or am I asking for too much since this construction, the determinant $d(t)$, is a bit arbitrary?

Associated Mathematica program:

(*Mathematica 8 start*)
nnn = 17;
Monitor[g = Table[Clear[s, x, t, nn, n, k];
   nn = ii;
   A = Table[
     Table[If[Mod[n, k] == 0, N[k/n^(1/4 (3 + 2 I t))], 0], {k, 1, 
       nn}], {n, 1, nn}];
   B = Table[
     Table[If[Mod[n, k] == 0, MoebiusMu[k]/n^N[(1/4 (3 + 2 I t))], 
       0], {k, 1, nn}], {n, 1, nn}];
   d = Det[A.Transpose[B] + IdentityMatrix[nnn]];
   Show[Plot[Re[d], {t, 0, 60}],
    Graphics[
     Table[Line[{{Im[ZetaZero[n]], -2}, {Im[ZetaZero[n]], 2}}], {n, 1,
        12}]], ImageSize -> Large], {ii, nnn, nnn}], ii]
g2 = Plot[
   2.3*Re[Sum[MoebiusMu[n]/n^(1/2 + I*t), {n, 1, nnn}]], {t, 0, 60}, 
   PlotStyle -> Red];
Show[g[[1]], g2]
(*Mathematica 8 end*)