I am just starting out in learning Linear Algebra and I have been trying to figure out something that has me confused. Per the rules, as I understand them, when reducing a row to obtain a matrices Row Echelon form you can:
1. Scale - Multiply all entries in a row by a nonzero number
2. Replace - Add a multiple of one row to another, replacing the second row with the result
3. Swap - Interchange two rows
This seems to make sense, however when I go to utilize this to solve for the inverse of a given 3x3 matrix A utilizing the identity matrix, where:
A = |40 7 29 | 1 0 0 |
| 0 11 75 | 0 1 0 |
| 5 25 27 | 0 0 1 |
The way I would think to start solving the problem would be is to take Row3 and perform the following to obtain a 0 in row 3, column 1:
A = |40 7 29 | 1 0 0 |
| 0 11 75 | 0 1 0 |
| 5 25 27 | 0 0 1 | R3: 8(R3) - R1
A = |40 7 29 | 1 0 0 |
| 0 11 75 | 0 1 0 |
| 0 193 187 | -1 0 1 |
However, when I have worked this out further, it gets to be very large and unmanageable with the large values in row 3 now... I have seen it solved elsewhere on symbolab and their solution in step 1 was instead to perform:
A = |40 7 29 | 1 0 0 |
| 0 11 75 | 0 1 0 |
| 5 25 27 | 0 0 1 | R3: R3 - 1/8(R1)
A = |40 7 29 | 1 0 0 |
| 0 11 75 | 0 1 0 |
| 0 193/8 187/8 | -1/8 0 1 |
I guess my main question is, how do you know how to solve these and which ways to go about it? Was my first solution never going to work because of how large things were getting or would it eventually have been correct, just more tedious? It just seems like I am supposed to intuitively know the magic number to scale, replace or swap with each time and if I choose incorrect it's a nightmare. Are there other rules that I am missing?
Thanks!