In the code that you have shared, the running times of each of the loops will get multiplied to each other.
We can visualise it something like this:
The loop iterating "k" will run "n" times for every iteration of "j". The loop iterating over "j" will run "i-1" times, as it runs from 2 to "i". Now, as "i" varies, we can see the loop iterating over "i" will run (n-1)*n. "i" itself has n iterations, so all the loops will run n*(n*(n-1)) ~ n^3.
The only thing I would like to point out that the line 2, will have a O(n^2) complexity, as allocation of space to a 2D matrix takes up time.
The overall complexity of the code will still be O(n^3).
Hope this helps.
