Decompostion with loss of information shown by junction
The trick to solve this exercise it to repeat as much as possible the elements of the table. Since A is the key, you cannot repeat A, but you can, for instance, repeat two times B and C and three times D. Here is a very simple and short example (note that the functional dependencies are respected):

r =
+---+---+---+---+
| A | B | C | D |
+---+---+---+---+
|a1 |b1 |c1 |d1 |
|a2 |b1 |c1 |d1 |
|a3 |b2 |c2 |d1 |
+---+---+---+---+


And since this is an exercise, I will leave to you the task of projecting r over R1, R2 and R3, and discovering how the natural join of the three relations differs from the original relation r.

Finally a terminological note: a relation instance is not an array! There is no order in the rows or the columns of a relation, while in an array the order is essential.