## Elementary Linear Algebra 7th Edition

The first row remained unchanged. The second row was changed: The first row multiplied by $3$ was added to it.
Here the first row is unchanged and the second row is modified. $$\text{Row 1 Original} = \left[\begin{matrix}3&-1&-4\end{matrix}\right]\Longrightarrow \text{Row 1 Transformed} = \left[\begin{matrix}3&-1&-4\end{matrix}\right];$$ $$\text{Row 2 Original} = \left[\begin{matrix}-4&3&7\end{matrix}\right]\Longrightarrow \text{Row 2 Transformed} = \left[\begin{matrix}5&0&-5\end{matrix}\right].$$ To transform a row we can only multiply each element in it by some number and add the element in the same column of another row multiplied by some number to it. To find out what was done here and be sure about it we will make a system of equations using the previous statement. So we need to have: \begin{align*} -4x+3y=&5\\ 3x-y=&0. \end{align*} It is enough to use only two elements and we picked 1st and 2nd. The third has to agree with our results for this to be a valid elementary row transformation. $x$ represents a number by which we multiplied the first row and $y$ is a number by which the second row is multiplied and added to the first one. Now lets solve this system: Step 1: Express $y$ from the second equation in terms of $x$ $$y=3x$$ Step 2: Put this into the first equation $$-4x+3\times3x=5\Rightarrow 5x=5$$ which gives $x=1$ this means that the second row wasn't premultiplied (multiplication by one changes nothing) Step 3: Use this to calculate $y$: $$y=3x=3\times 1 = 3.$$ This means that the second row was multiplied by $3$ and then added to the first.