It's a shame you feel that way. Modifying an element of an array in Haskell is easy and the blog post gives almost exactly correct code, but for some reason places it under the title "we run into trouble when we realize there’s no built-in mutating assignment". But we don't run into trouble! Mutable arrays are provided in Haskell! Haskell has mutation! Here's the code

    import Data.Array.MArray
    import Data.Array.IO
    
    size = 10
    
    (!) = readArray
    
    main :: IO ()
    main = do
      -- 1. Declare the array
      arr <- newArray ((1,1), (size,size)) undefined
      let _ = arr :: IOArray (Int,Int) Integer
    
      -- 2. Initialize the array to 0
      sequence_ $ do
        i <- [1..size]
        j <- [1..size]
        return $ writeArray arr (i, j) 0
    
      -- 3. Set the diagonal to 1
      sequence_ $ do
        i <- [1..size]
        return $ writeArray arr (i, i) 1
    
      -- 4. Print the array
      sequence_ $ do
        i <- [1..size]
        j <- [1..size]
        return $ do
          arr_i_j <- arr ! (i,j)
    
          putChar $ if arr_i_j == 0
                    then '0'
                    else '1'
          if j == size
            then putChar '\n'
            else return () 

    > main
    1000000000
    0100000000
    0010000000
    0001000000
    0000100000
    0000010000
    0000001000
    0000000100
    0000000010
    0000000001

I've no idea why the blog post is written in such long winded style. Mutation in Haskell is not difficult!