Linear Algebra 9g: Which Columns Are Pivot Columns in Gaussian Elimination

[pendjme]

Thought I had this previously and was wrong. Key for me is that the linearly dependent columns ( col 2 and 4 ) are made up of the two independent columns ( col 1 and 3) that come before it.
Was confusing to me how you knew that column 2 was 3 times column 1. Could see this was true, but it took a minute for me to see it was made up of the preceding independent column.
Same for column 4 being made up of the two preceding independent columns. Was not clear to me why column 2 was left out of column 4 until realized that 2 was dependent on col 1.

Thanks again for this super course.

[TheMrDJD]

All of this made sense, but at 8:54 things started to get confusing. "One of this column and one of the third". That kind of made sense as it means whatever the values are for x and z are linearly independent. Even "the second column is 3 times the first" kind of makes sense. We are using a scalar, alpha, because of the dependence, right? The final beta scalar is then trying to account for the second dependent column, but that's a little harder to piece together.

[ridwanwase]

What if i switch column and RHS same time accordingly then do our different arrangement bring any change in which columns will get pivot element?

[abajabbajew]

Should some way be made to emphasize the pivots in positions '1' and '3' ('x' and 'z') in the particular solution vector together with the _complementary_ positions of the other - I suppose 'slack' - variables of '-1' in positions '2' ('y') and '4' ('t') in each of the _null_ vectors? Since those particular entries will always be '-1', perhaps putting a these in brackets (i.e. as '(-1)' ) within their columns, suggesting complements, might be appropriate to bring out the relations?


|  1 |   |  3 |   | -1 |
|  0 |   |(-1)|   |  0 |
|  1 |   |  0 |   |  2 |
|  0 |   |  0 |   |(-1)|

[NTC]

4:18, why did you  eliminate  the entry above the pivot, instead of eliminating the first entry of the last column ?