(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given matrix

A=

1 1 -2 1 3

2 -1 2 2 6

3 2 -4 -3 -9

x=

x1

x2

x3

x4

x5

b =

1

2

3

1. Solve the system by elimination (use the augmented matrix) until pivots are found (no backward elimination here).

2. Show that pivots are respectively 1 and -3, and indicate clearly which are the pivot columns and free columns.

2. Relevant equations

Matrix A and vector b

3. The attempt at a solution

I perform elimination on augmented matrix to solve part 1.

1 1 -2 1 3 | 1

0 -3 6 0 0 | 0

0 -1 2 -6 -18 | 0

1 1 -2 1 3 | 1

0 -3 6 0 0 | 0

0 0 0 -6 -18 | 0

What I don't understand is part 2.

As far as I know, for an entry to become a pivot it must be the first non-zero entry on the row and there must be no non-zero entries under it.

1 1 -2 1 3 | 1

0 -3 6 0 0 | 0

0 0 0 -6 -18 | 0

If that rule applies I would find three pivots instead of two. (1,-3, -6) I know the right answer for this exercise is that there are only two pivots (1, -3), resulting col1 & col2 to be the pivot columns and the rest are free columns.

Can someone explain why -6 is not a pivot in this case?

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# Which are the pivot columns? - Linear Algebra

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