Gauss Jordan Elimination. How to know where to start?

[hobomath]

Hello, I'm having trouble with the Gauss Jordan elimination method to find the inverse of a 3x3 Matrix.
I understand how it works ( [A | I ] to [ I | A^-1] ), but I was wondering if there was a trick in knowing what number to tackle next?

For exemple with this matrice:
0 -2 2
1 1 0
2 0 -3

I'll start by finding the '1' on the top left by swapping R₁ with R₂.
1 1 0
0 -2 2
2 0 -3

Cool. I got the first number. Now what? Do I solve each number clockwise? Anti-clockwise? By column? The '1's first? Or is it really by trial and error?
In that case, how do I know that the matrix is not invertible? :/

Thank you for helping.
Edit: for clarification, I can solve the matrix. Was wondering if there was an optimal route to solve it + know when it's not possible to find the inverse of it.

 

[stapel]

Hello, I'm having trouble with the Gauss Jordan elimination method to find the inverse of a 3x3 Matrix.
I understand how it works ( [A | I ] to [ I | A^-1] ), but I was wondering if there was a trick in knowing what number to tackle next?

There is, to my knowledge, no hard and fast rule. You pick the steps, in accordance with what you might personally feel useful. It is entirely possible for you and your friends each to take a different series of steps, and yet end up with the same (correct) solution at the end.

For exemple with this matrice:
0 -2 2
1 1 0
2 0 -3

I'll start by finding the '1' on the top left by swapping R₁ with R₂.
1 1 0
0 -2 2
2 0 -3

Cool. I got the first number. Now what?

You know that you want to end up with the left-most columns as simplified as possible (that is, all 1's and 0's, with as many 0's as you can manage), so do row operations until you get this.

One student might prefer to multiply R3 by -1/2 next, followed by adding R1 to R3 to clear out the first entry in R3. Me, I prefer to avoid fractions as long as possible, so I'd rather Add -2*R1 to R3:

-2*R1: -2 -2 0

-2*R1 + R3 = new R3: 0 -2 -3

This would give me the following:

1 1 0
0 -2 2
0 -2 -3

My preference then would be to add -1*R3 to R2, giving me:

1 1 0
0 0 1
0 -2 -3

And so forth. But somebody else might have a different preference. As long as each step were mathematically valid, then the results will be valid. As long as the steps lead logically toward the desired result, the final matrix should be the same (after swapping rows, perhaps). There is no one and only "right" sequence of steps.

 

[Ishuda]

Hello, I'm having trouble with the Gauss Jordan elimination method to find the inverse of a 3x3 Matrix.
I understand how it works ( [A | I ] to [ I | A^-1] ), but I was wondering if there was a trick in knowing what number to tackle next?

For exemple with this matrice:
0 -2 2
1 1 0
2 0 -3

I'll start by finding the '1' on the top left by swapping R₁ with R₂.
1 1 0
0 -2 2
2 0 -3

Cool. I got the first number. Now what? Do I solve each number clockwise? Anti-clockwise? By column? The '1's first? Or is it really by trial and error?
In that case, how do I know that the matrix is not invertible? :/

Thank you for helping.
Edit: for clarification, I can solve the matrix. Was wondering if there was an optimal route to solve it + know when it's not possible to find the inverse of it.

I've never even tried to solve any (nearly full) matrix over a 3X3 by hand [well maybe once or twice]. The method I use is the augmented matrix approach which I assume you also use. I work on element (1,1) to make it 1, then make the rest of column 1 zero [subtract element (1,n) times row 1 from row n]. Now make (2,2) equal to 1 and then make the rest of column 2 zero [subtract element (2,n) times row 2 from row n. Because column 1 is zero except for (1,1) column 1 will remain unchanged]. Repeat for column 3, making element (3,3) equal to 1 and the rest of column 3 equal to zero. If you have larger than a 3X3 repeat ad nausium until you have the identity matrix. The augmented matrix will be the inverse [or the solution to the system of equations].

Simple example for 2X2: Find the inverse of \(\displaystyle \begin{pmatrix}0& 1\\ 1& 4 \end{pmatrix}\)
\(\displaystyle \begin{pmatrix}0& 1& 1& 0\\ 1& 4& 0& 1 \end{pmatrix}\)
interchange rows 1 and 2
\(\displaystyle \begin{pmatrix}1& 4& 0& 1\\ 0& 1& 1& 0 \end{pmatrix}\)
Element (1,1) is 1 and the rest of column 1 is zero, so move on to column 2. Element (2,2) is 1 so nothing to do there. But we need to make element (1,2) zero so subtract 4 times row 2 from row one
\(\displaystyle \begin{pmatrix}1& 0& -4& 1\\ 0& 1& 1& 0 \end{pmatrix}\)

So the inverse is \(\displaystyle \begin{pmatrix}-4& 1\\ 1& 0 \end{pmatrix}\)

Of course, if you see short cuts, take them. Like you seeing a 1 in column 1 and interchanging rows so that you now have 1 for element (1,1)

 

[hobomath]

Alrighty, thank you stapel!
I appreciate the information! That what I wanted to know!