code: function linearDependencies(matrix, options = {}) {
                const { thresholdValue = 10e-10, thresholdError = 10e-10 } = options;
  
                var n = matrix.rows;
                var results = new Matrix$1(n, n);
  
                for (var i = 0; i < n; i++) {
                  var b = Matrix$1.columnVector(matrix.getRow(i));
                  var Abis = matrix.subMatrixRow(xrange(n, i)).transposeView();
                  var svd = new SingularValueDecomposition$$1(Abis);
                  var x = svd.solve(b);
                  var error = max(
                    Matrix$1.sub(b, Abis.mmul(x))
                      .abs()
                      .to1DArray()
                  );
                  results.setRow(
                    i,
                    dependenciesOneRow(error, x, i, thresholdValue, thresholdError)
                  );
                }
                return results;
              }
docs: Creates a matrix which represents the dependencies between rows.
  If a row is a linear combination of others rows, the result will be a row with the coefficients of this combination.
  For example : for A = [[2, 0, 0, 1], [0, 1, 6, 0], [0, 3, 0, 1], [0, 0, 1, 0], [0, 1, 2, 0]], the result will be [[0, 0, 0, 0, 0], [0, 0, 0, 4, 1], [0, 0, 0, 0, 0], [0, 0.25, 0, 0, -0.25], [0, 1, 0, -4, 0]]
  @param {Matrix} matrix
  @param {Object} [options] includes thresholdValue and thresholdError.
  @param {number} [options.thresholdValue = 10e-10] If an absolute value is inferior to this threshold, it will equals zero.
  @param {number} [options.thresholdError = 10e-10] If the error is inferior to that threshold, the linear combination found is accepted and the row is dependent from other rows.
  @return {Matrix} the matrix which represents the dependencies between rows.