Question

Hi guys, small question. When he talks about the pivots in lecture 7, minute 7:30 approximately, why does he choose this two as points. I would have rather taken the position A(2,2) rather than A(2, 3) as he takes the second pivot. Why is that so?

Thanks!

Answer 1

First, pivots can not be zero.
Second, the 2nd column is all zeros (except for the first entry), so there is no "work" to be done on that column

Answer 2

The second column is [2 0 0]^T. I agree that pivots cannot be zero. However, what I am asking is, what is the reason why he chooses the pivot in row 2, but column 3 rather than in column 2.
What I can think of is that this column is linearly dependent, so it is like if it was not there. Is that it?

Thank you

Answer 3

Perhaps I don't understand your question.

There is only one pivot per row, so we can't use entry (1,2) as the next pivot, which must be in row 2.

Entry (2,2) is zero, so we can't choose it as the pivot for the 2nd row.
If there were non-zeros below (2,2) we could swap rows, and then choose the entry at (2,2). But all other entries in col 2 are zero, so we have to move to the 3rd entry in row 2,
and use the non-zero entry at (2,3) as the pivot.


yes, the 2nd column is a multiple of the first column. This shows column 2 is a linear combination of the "pivot columns" that come before. (the combination is given by the entries in the column... for example the 2 in (1,2) means 2 * the first column)