g | x | w | all
Bytes Lang Time Link
049JavaScript Node.js241021T013855Zl4m2
010Uiua241021T012440Znyxbird
086R201120T153121ZGiuseppe
010APL Dyalog Unicode201119T193953Zfireflam
037Julia160428T160748ZDennis
079JavaScript ES6160428T125208Zedc65
010J160428T220830Zmiles
009Jelly160428T124017ZLeaky Nu
042MATLAB / Octave160428T131425ZStewie G
011Pyth160428T125200ZLeaky Nu
013CJam160428T123115ZMartin E

JavaScript (Node.js), 49 bytes

f=(a,b)=>a.flatMap(t=>b.map(u=>t.map?f(t,u):t*u))

Try it online!

Nothing special. Can save extra byte if [].at exist

Uiua, 10 bytes

/⊂≡/≡⊂פ¤:

Try it!

I wouldn't be surprised if this was suboptimal. If we can take input reversed, remove : flip for -1 byte. ¤ fixing twice forces × multiply to vectorize, and then we / reduce ⊂ join across various ≡ rows.

R, 98 86 bytes

function(x,y,`+`=array)aperm(apply(x,1:2,`*`,y)+c(w<-dim(y),v<-dim(x)),c(1,3,2,4))+v*w

Try it online!

Reimplementation of .kronecker and outer for matrices. I do think there's a golfier approach out there, maybe using apply? 6 bytes golfed using apply and array thanks to Dominic van Essen!

The builtins are %x% for kronecker(A,B,"*") and %o% for outer(A,B,"*").

R, 120 bytes

function(A,B){l=list()
a=dim(A)
for(i in 1:a[2]-1)l[[i+1]]<-do.call(rbind,lapply(A,"*",B)[1:a+a[2]*i])
do.call(cbind,l)}

Try it online!

Naive approach: calculate the subarrays \$a_{ij}B\$ and bind them together in the appropriate order. There are probably golfs here too, but I don't think it'll be shorter than the one above.

APL (Dyalog Unicode), 10 bytes

{⍪/,/⍺×⊂⍵}

Try it online!

How it Works

{  ...   }  ⍝ dfn that takes ⍺=A and ⍵=B
     ⍺×⊂⍵   ⍝ Product of each element of A with all of matrix B
            ⍝ Gives a nested array: an matrix of matrices
   ./       ⍝ Join rows
 ⍪/         ⍝ Join columns

Julia, 40 39 37 bytes

A%B=hvcat(sum(A^0),map(a->a*B,A')...)

Try it online!

How it works

JavaScript (ES6), 79

Straightforward implementation with nested looping

(a,b)=>a.map(a=>b.map(b=>a.map(y=>b.map(x=>r.push(y*x)),t.push(r=[]))),t=[])&&t

Test

f=(a,b)=>a.map(a=>b.map(b=>a.map(y=>b.map(x=>r.push(y*x)),t.push(r=[]))),t=[])&&t

console.log=x=>O.textContent+=x+'\n'

function show(label, mat)
{
  console.log(label)
  console.log(mat.join`\n`)
}

;[ 
  {a:[[1,2],[3,4]],b:[[5,6],[7,8]] },
  {a:[[1],[2]],b:[[1,2]]},
  {a:[[16,2,3,13],[5,11,10,8],[9,7,6,12],[4,14,15,1]],b:[[1,1],[0,1]]},
  {a:[[2]],b:[[5]]}
].forEach(t=>{
  show('A',t.a)  
  show('B',t.b)
  show('A⊗B',f(t.a,t.b))
  show('B⊗A',f(t.b,t.a))  
  console.log('-----------------')
})
<pre id=O></pre>

J, 10 bytes

This is one possible implementation.

[:,./^:2*/

J, 13 bytes

This is a similar implementation, but instead uses J's ability to define ranks. It applies * between each element on the LHS with the entire RHS.

[:,./^:2*"0 _

Usage

   f =: <either definition>
    (2 2 $ 1 2 3 4) f (2 2 $ 5 6 7 8)
 5  6 10 12
 7  8 14 16
15 18 20 24
21 24 28 32
   (2 1 $ 1 2) f (1 2 $ 1 2)
1 2
2 4
   2 f 5
10

Jelly, 10 9 bytes

×€€;"/€;/

Uses Büttner's Algorithm (ü pronounced when trying to make an ee sound [as in meet] in the mouth-shape of an oo sound [as in boot]).

The ;"/€;/ is inspired by Dennis Mitchell. It was originally Z€F€€;/ (which costs one more byte).

MATLAB / Octave, 83 42 Bytes

Saved 41 bytes, thanks to FryAmTheEggman!

@(A,B)cell2mat(arrayfun(@(n)n*B,A,'un',0))

Test it here!

Breakdown

arrayfun is a disguised for-loop that multiplies n*B, for a variable n defined by the second argument. This works because looping through a 2D matrix is the same as looping through a vector. I.e. for x = A is the same as for x = A(:).

'un',0 is equivalent to the more verbose 'UniformOutput', False, and specifies that the output contains cells instead of scalars.

cell2mat is used to convert the cells back to a numeric matrix, which is then outputted.

Pyth, 14 12 11 bytes

JEsMs*RRRRJ

Translation of Jelly answer, which is based on Büttner's Algorithm (ü pronounced when trying to make an ee sound [as in meet] in the mouth-shape of an oo sound [as in boot]).

Try it online (test case 1)!

Bonus: calculate B⊗A in the same number of bytes

JEsMs*LRLRJ

Try it online (test case 1)!

CJam, 13 bytes

{ffff*::.+:~}

This is an unnamed block that expects two matrices on top of the stack and leaves their Kronecker product in their place.

Test suite.

Explanation

This is just the Kronecker product part from the previous answer, therefore I'm here just reproducing the relevant parts of the previous explanation:

Here is a quick overview of CJam's infix operators for list manipulation:

ffff*  e# This is the important step for the Kronecker product (but
       e# not the whole story). It's an operator which takes two matrices
       e# and replaces each cell of the first matrix with the second matrix
       e# multiplied by that cell (so yeah, we'll end up with a 4D list of
       e# matrices nested inside a matrix).
       e# Now the ffff* is essentially a 4D version of the standard ff* idiom
       e# for outer products. For an explanation of ff*, see the answer to
       e# to the Kronecker sum challenge.
       e# The first ff maps over the cells of the first matrix, passing in the 
       e# second matrix as an additional argument. The second ff then maps over 
       e# the second matrix, passing in the cell from the outer map. We 
       e# multiply them with *.
       e# Just to recap, we've essentially got the Kronecker product on the
       e# stack now, but it's still a 4D list not a 2D list.
       e# The four dimensions are:
       e#   1. Columns of the outer matrix.
       e#   2. Rows of the outer matrix.
       e#   3. Columns of the submatrices.
       e#   4. Rows of the submatrices.
       e# We need to unravel that into a plain 2D matrix.
::.+   e# This joins the rows of submatrices across columns of the outer matrix.
       e# It might be easiest to read this from the right:
       e#   +    Takes two rows and concatenates them.
       e#   .+   Takes two matrices and concatenates corresponding rows.
       e#   :.+  Takes a list of matrices and folds .+ over them, thereby
       e#        concatenating the corresponding rows of all matrices.
       e#   ::.+ Maps this fold operation over the rows of the outer matrix.
       e# We're almost done now, we just need to flatten the outer-most level
       e# in order to get rid of the distinction of rows of the outer matrix.
:~     e# We do this by mapping ~ over those rows, which simply unwraps them.