I'm trying to learn about solving equation systems using the Gauss-Jordan method. So, you have to convert the equation system to a matrix, and then reduce it to the identity. When you transform it to the augmented matrix, it is pretty easy to find the solutions by going backwards...
- Can all equation systems be reduced to the identity matrix? If not, how do you determine whether you can or not?
And, as a side note:
- From my course, it would seem like this is pretty much a "lightbulb" technique. You literally stare at the matrix and suddenly it occurs to you how to proceed. Is that correct? Is there really no standard?
Not all systems of linear equations can be reduced to the identity matrix. A (square) matrix can only be reduced to the identity if the matrix is invertible (i.e., has an inverse). There are many ways to check if a matrix is invertible; some ways include checking if the determinant is non-zero, or if for an $n\times n$ matrix, the rank is $n$.
I wouldn't really say the Gauss-Jordan algorithm is a "lightbulb" technique. It's an algorithm, so you just keep repeating it until the matrix is in (Reduced) Row Echelon Form.
Given for example, \begin{bmatrix} 2&4&2\\ 0&2&3\\ 3&2&1 \end{bmatrix} You start with the $a_{11}=2$ and make it $1$ by multiplying the first row by $\frac{1}{2}$.
$\to \begin{bmatrix} 1&2&1\\ 0&2&3\\ 3&2&1 \end{bmatrix}$
Then you want to make sure $a_{21}=0$. It is already $0$, so we are in luck. So let's move on. We want $a_{31}=1$ next; it's a $3$, so let's multiply row $1$ by $3$ and subtract row $3$.
$\to \begin{bmatrix} 1&2&1\\0&2&3\\0&4&2 \end{bmatrix}$.
Then you want to make $a_{22}=1$ by multiplying row $2$ by $\frac{1}{2}$. And you continue this procedure.
Essentially, you multiply the row by some constant in order to make the term you want into a $1$, and then multiply that $1$ by some constant and add or subtract some row until the numbers below and above it are $0$, just like how we did it for the first column.