Java:

• method name should be camelCase, not PascalCase

Refer here on how to enable backward compatability to not invalidate all solutions

All languages:

• The output can be easily hard-coded since for each n, there is only 1 valid upper / lower triangular matrix. Hence, the input range should be increased (Also, it should cover range from 1 - 3)

Ruby:

• Attempts tests don't include fixed tests.
• Empty matrix is not tested.

• Node 18 should be enabled (Refer this and this for more info)

• Ruby 3.0 should be enabled (Refer this & this for more detail)

A pretty straightforward algorithm to calculate, and I was able to make a reasonably efficient one, I think...

I feel like the description could use a little work. A few specific points:

• You never say "top-right" or "bottom-left" in your description, which is pretty necessary information if the reader's going to actually produce something along the right lines.
• You should also say if the diagonal itself (items along that line) should be 0 or not.
• The examples are useful, but it's hard to see how they actually form a grid. Perhaps something like the below would be easier to visualise?

upperTriangular([
  [1,1,1],
  [0,1,1],
  [0,0,1]
]) == true