What is the best way to get matrices into Row-Eschelon form?

Question

I'm having a very hard time understanding this. I see many examples done, but when I do it it takes forever and is very frustrating since I screw up all the time and get wrong answers etc.

What is the best / fastest / easiest way to think about this. Perhaps there is some kind of trick to figure out how to do it?

Answer 1

It's hard to put in words what the best way is. The key thing to do is *make* leading ones, and then give it surroundings zeros by performing the appropriate operations. Here's an example below.

[2 -2 4]
[-1 2 -1]
[4 1 0]

R1 -> (1/2) R1

[1 -1 2]
[-1 2 -1]
[4 1 0]

R2 -> R2 + R1
R3 -> R3 - 4R1

[1 -1 2]
[0 3 -1]
[0 5 -8]

Now, start with the second row.

R2 -> (1/3)R2

[1 -1 2]
[0 1 -1/3]
[0 5 -8]

R1 -> R1 + R2
R3 -> R3 - 5R2

[1 0 5/3]
[0 1 -1/3]
[0 0 -19/3]

R3 -> -3/19 R3

[1 0 5/3]
[0 1 -1/3]
[0 0 1]

R1 -> R1 - 5/3 R3
R2 -> R2 + (1/3) R3

[1 0 0]
[0 1 0]
[0 0 1]

******
In some cases, making what's supposed to be the leading 1 is not possible, such as this one:

[0 2 0]
[4 4 4]
[1 3 2]

If that's the case, your very first step is to SWAP rows. In this case, it would be convenient to swap the 1st row with the 3rd row.

R1 <--> R3

[1 3 2]
[4 4 4]
[0 2 0]

And then attempt to solve as normal.

To answer your questions, it's all about knowing how to create leading 1s and surrounding those 1s with 0s.