Use jldoctest in docstrings (#72)

src/frobenius.jl

@@ -55,7 +55,7 @@ julia> F*F # Special form preserved if the same column has the subdiagonals
  0  0  12  0  0  1
 
 julia> F*Frobenius(2, 2:5) # Promotes to Matrix
-6×6 Matrix{Float64}:
+6×6 Matrix{Int64}:
  1   0  0  0  0  0
  0   1  0  0  0  0
  0   2  1  0  0  0

src/hilbert.jl

@@ -4,33 +4,36 @@
 export Hilbert, InverseHilbert
 
 """
-[`Hilbert` matrix](http://en.wikipedia.org/wiki/Hilbert_matrix)
+   Hilbert(m [,n])
 
-```julia
-julia> A=Hilbert(5)
-Hilbert{Rational{Int64}}(5,5)
+Construct `m × m` or `m × n`
+[`Hilbert` matrix](http://en.wikipedia.org/wiki/Hilbert_matrix)
+from its specified dimensions,
+where element `i,j` equal to `1 / (i+j-1)`.
 
-julia> Matrix(A)
-5x5 Array{Rational{Int64},2}:
+```jldoctest hilbert1
+julia> A = Hilbert(5)
+5×5 Hilbert{Rational{Int64}}:
  1//1  1//2  1//3  1//4  1//5
  1//2  1//3  1//4  1//5  1//6
  1//3  1//4  1//5  1//6  1//7
  1//4  1//5  1//6  1//7  1//8
  1//5  1//6  1//7  1//8  1//9
 
-julia> Matrix(Hilbert(5))
-5x5 Array{Rational{Int64},2}:
+julia> Matrix(A)
+5×5 Matrix{Rational{Int64}}:
  1//1  1//2  1//3  1//4  1//5
  1//2  1//3  1//4  1//5  1//6
  1//3  1//4  1//5  1//6  1//7
  1//4  1//5  1//6  1//7  1//8
  1//5  1//6  1//7  1//8  1//9
 ```
+
 Inverses are also integer matrices:
 
-```julia
+```jldoctest hilbert1
 julia> inv(A)
-5x5 Array{Rational{Int64},2}:
+5×5 InverseHilbert{Rational{Int64}}:
     25//1    -300//1     1050//1    -1400//1     630//1
   -300//1    4800//1   -18900//1    26880//1  -12600//1
   1050//1  -18900//1    79380//1  -117600//1   56700//1

src/kahan.jl

@@ -5,25 +5,25 @@ export Kahan
 """
 [`Kahan` matrix](http://math.nist.gov/MatrixMarket/data/MMDELI/kahan/kahan.html)
 
-```julia
-julia> A=Kahan(5,5,1,35)
-5×5 Kahan{Int64,Int64}:
+```jldoctest
+julia> A = Kahan(5,5,1,35)
+5×5 Kahan{Int64, Int64}:
  1.0  -0.540302  -0.540302  -0.540302  -0.540302
  0.0   0.841471  -0.454649  -0.454649  -0.454649
  0.0   0.0        0.708073  -0.382574  -0.382574
  0.0   0.0        0.0        0.595823  -0.321925
  0.0   0.0        0.0        0.0        0.501368
 
-julia> A=Kahan(5,3,0.5,0)
-5×3 Kahan{Float64,Int64}:
+julia> A = Kahan(5,3,0.5,0)
+5×3 Kahan{Float64, Int64}:
  1.0  -0.877583  -0.877583
  0.0   0.479426  -0.420735
  0.0   0.0        0.229849
  0.0   0.0        0.0
  0.0   0.0        0.0
 
-julia> A=Kahan(3,5,0.5,1e-3)
-3×5 Kahan{Float64,Float64}:
+julia> A = Kahan(3,5,0.5,1e-3)
+3×5 Kahan{Float64, Float64}:
  1.0  -0.877583  -0.877583  -0.877583  -0.877583
  0.0   0.479426  -0.420735  -0.420735  -0.420735
  0.0   0.0        0.229849  -0.201711  -0.201711