Sorry if this is very simple, I just could not find a good explanation for this particular hangup on Gaussian elimination online.
Let's say that I am trying to solve a system of linear equations using a 3×3 matrix (let's call it $A$ ) and a vector $r$ corresponding to the equation's output values.
$A\\\begin{bmatrix}1&1&3\\1&2&4\\1&1&2\end{bmatrix}$ $s\\\begin{bmatrix}a\\b\\c\end{bmatrix}$ $r\\\begin{bmatrix}15\\21\\13\end{bmatrix}$
One method to do this is by first using Gaussian elimination to convert $A$ into echelon form, then use back substitution to find the values of $a$,$b$, and $c$.
When using reduction, we can add a row of $A, s, r$ onto another row, or multiply a row by a non 0 constant. My instructor has said that this "doesn't change" change the value of the system of equations. E.g.:
subtracting row 1 from row 3
$A\\\begin{bmatrix}1&1&3\\1&2&4\\0&0&-1\end{bmatrix}$ $s\\\begin{bmatrix}a\\b\\c\end{bmatrix}$ $r\\\begin{bmatrix}15\\21\\-2\end{bmatrix}$
multiplying row 3 by -1
$A\\\begin{bmatrix}1&1&3\\1&2&4\\0&0&1\end{bmatrix}$ $s\\\begin{bmatrix}a\\b\\c\end{bmatrix}$ $r\\\begin{bmatrix}15\\21\\2\end{bmatrix}$
How is it that we're not changing any values here? I understand if we're doing an operation on both sides of an equation it would still be true because the $=$ would still hold, but with matrices and vectors there's no = operator. It seems like we're basically making arbitrary changes to the matrix, which would scale the vector differently.
Best Answer
The reason it is valid here is because your $r$ vector is what would be on the other side of the equals sign. You've put it into a form that allows you to use a method that you understand (row reduction) to get your result. The $r$ actually does belong on the other side of the equals sign, because then you can perform matrix multiplication between $A$ and $s$ to get a 3x1 matrix, exactly like $r$, and then you can compare terms. I think the idea here is you've put it in this form simply because it's easier for you to row reduce this way.