Problem with the Rank of a matrix after a FindRoot

Question

I am solving a boundary problem defined by a set of four-component vectorial functions. Without going into many details, I have two functions on the left and two on the right of an interface. The boundary problem is defined by requiring that the sum of the two functions on the left is equal to the sum of the two on the right at the interface position. The functions contain confluent hypergeometric functions of the first and second kind.

A similar problem is studied in this work but with two-components functions.

The problem admits a unique solution when the determinant of the matrix associated is zero. There is a free parameter $j$, and for a given value of it, I can determine with FindRoot the associated value of the energy $e$. For this pair of number, the determinant is smaller than $10^{-10}$. However, the rank of the associated matrix is not reduced to 3 but is still 4. To reduce the Rank of the matrix, I need to work around the solution $e$ until the determinant is not smaller than $10^{-16}$!

The two vectorial functions are

WFmFR[j_, e_, b_, V0_, x_, lR_] := Module[{wf, av, mm, mp,U,M},
  M[a_, b_,x_] := Hypergeometric1F1[a, b, x] 
  U[a_, b_,x_] := HypergeometricU[a, b, x]
  mm = (e - V0)^2;
  mp = e - V0;
  av = UnitStep[j] j - 
    1/2 (+(e - V0)^2 - 1 + Sqrt[1 + lR^2 (e - V0)^2] b);
  wf = E^(-x/2)
     If[j > 0, 
     x^(j/2) {Sqrt[2] j x^(-1/2) lR  M[av, j, x], 
       lR  mp M[av, j + 1, x], -(2 (av - j) + mm) M[av, j + 1, x], 
       Sqrt[2] ((av - j - 1) (2 (av - j) + mm))/(mp (j + 1)) x^(1/2)
         M[av, j + 2, x]}, 
     If[j == 0, {Sqrt[2] av x^(1/2) lR  M[av + 1, 2, x], 
       lR  mp M[av, 1, x], -(2 av + mm) M[av, 1, x], 
       Sqrt[2]/mp (av - 1) (2 av + mm) x^(1/2) M[av, 2, x]}, 
      x^(-j/2) {(Sqrt[2] av)/(1 - j) lR  x^(1/2) M[av + 1, -j + 2, x],
         lR  mp M[av, -j + 1, x], -(2 av + mm) M[av, -j + 1, 
          x], -Sqrt[2] (j (2 av + mm))/mp x^(-1/2)
          M[av - 1, -j, x]}]]; 
  Normalize[wf]]

and

WFMFR[j_, e_, b_, V0_, x_, lR_] := Module[{wf, av, mm, mp,M,U},
  M[a_, b_,x_] := Hypergeometric1F1[a, b, x] 
  U[a_, b_,x_] := HypergeometricU[a, b, x]
  mm = (e - V0)^2;
  mp = e - V0;
  av = UnitStep[j] j - 
    1/2 (+(e - V0)^2 - 1 + Sqrt[1 + lR^2 (e - V0)^2] b);
  wf = E^(-x/2)
     If[j > 0, 
     x^(j/2) {Sqrt[2] (j - av) x^(-1/2) lR  U[av, j, x], 
       lR  mp U[av, j + 1, x], -(2 (av - j) + mm) U[av, j + 1, x], -(
         Sqrt[2]/mp) (2 (av - j) + mm) x^(1/2) U[av, j + 2, x]}, 
     If[j == 0, {Sqrt[2] (-av) x^(1/2) lR  U[av + 1, 2, x], 
       lR  mp U[av, 1, x], -(2 av + mm) U[av, 1, x], -(Sqrt[2]/
         mp) (2 av + mm) x^(1/2) U[av, 2, x]}, 
      x^(-j/2) {-Sqrt[2] av x^(1/2) lR   U[av + 1, -j + 2, x], 
        lR  mp U[av, -j + 1, x], -(2 av + mm) U[av, -j + 1, x], -(
          Sqrt[2]/mp) (2 av + mm) x^(-1/2) U[av - 1, -j, x]}]];
  Normalize[wf]]

The matrix is defined as

MFR[j_, e_, rr_, V0_, lR_] := Module[{c1, c2, c3, c4, mat},
  c1 = WFmFR[j, e, +1, -V0, rr, lR];
  c2 = WFmFR[j, e, -1, -V0, rr, lR];
  c3 = -WFMFR[j, e, +1, V0, rr, lR];
  c4 = -WFMFR[j, e, -1, V0, rr, lR];
  mat = {c1, c2, c3, c4};
  mat]

For a fixed value of $j$, I am interested in the value of $e$ that makes zero of the determinant of this matrix. To find all zeros in a fixed interval, I used a routine suggested in the Wolfram Community Forum. I do not copy the function FindRoots since it can be copied from there.

To show my problem, I am evaluating a set of zeros in a specific interval

  FindRoots[Det[MFR[-2, x, 5., 0.50, 0.1]], {x, -5, 5}, 
    Debug -> True]

These are nullifying the determinant.

Det[MFR[-2, #, 5., 0.50, 0.1]] & /@ test

However, not all of the solutions are decreasing the rank of the matrix:

MatrixRank[MFR[-2, #, 5., 0.50, 0.1]] & /@ test