Back substitution is giving me error

Course Q&A Math for Machine Learning Linear Algebra for Machine Learning and Data Sc...
1

Hi team, the back substitution is giving me errors, I am not sure where I did wrong…
{mentor edit: code removed}

2

Please don’t post your code on the forum. That’s not allowed by the Code of Conduct.

Hint: Where you used “row-2”, that’s undoubtedly an error.

3

I tried row -1 is still the same, this is to get the third row if there are 4 rows right?

4

The error is
An exception was thrown while running your function: index 7 is out of bounds for axis 0 with size 3.
Input matrix:
[[1 0 0 5]
[0 1 0 6]
[0 0 1 7]]

5

NVM found the error,

The index = get_index_first_non_zero_value_from_row(M,row) I was passing the whole row array into the integer parameter.

Also I was misunderstand the process, on paper I normally work from bottom row’s non- zero column which is above a pivot to the top. In here we did it from r0 not r4 so we j=should use

        # mentor edit: code removed
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