## Elementary Linear Algebra 7th Edition

The second row multiplied by $4$ was added to the third row. The places of the first and the second row were exchanged.
Places of the first and second row were exchanged. The third row was modified $$\text{Row 1 Original} = \left[\begin{matrix}0&-1&-5&5\end{matrix}\right]\Longrightarrow \text{Row 1 Transformed} = \left[\begin{matrix}1&3&-7&6\end{matrix}\right];$$ $$\text{Row 2 Original} = \left[\begin{matrix}-1&3&-7&6\end{matrix}\right]\Longrightarrow \text{Row 2 Transformed} = \left[\begin{matrix}0&-1&-5&5\end{matrix}\right].$$ $$\text{Row 3 Original} = \left[\begin{matrix}4&-5&1&3\end{matrix}\right]\Longrightarrow \text{Row 3 Transformed} = \left[\begin{matrix}0&7&-27&27\end{matrix}\right].$$ 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*} -y+4z=&0\\ -x+3y-5z=&7\\ -5x-7y+z=&-27 \end{align*} It is enough to use only three elements and we picked 1st, 2nd and 3rd. The 4th 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 (pre exchange) and added it tho the third, $y$ is a number by which the second row (pre exchange) is multiplied and added to the third and $z$ is the number by which the third row was multiplied before all other transformations. Now lets solve this system: Step 1: Subtract from the third equation the second one multiplied by $5$ to eliminate $x$: $$-5x-5(-x) - 7y-5(3y)+z-5(-5z) = -27-5\times 7$$ which becomes $$-22y+26z=-62.$$ Divide this by $2$ for better looks: $$-11y+13z=-31.$$ Now our system reads \begin{align*} -y+4z=&0\\ -x+3y-5z=&7\\ -11y+13z=&-31 \end{align*} Step 2: Subtract from the third equation the first one multiplied by $11$: $$-11y-11(-y)+13z-11(4z)=-31-11\times0$$ which becomes $$-31z=-31$$ and that gives $$z=1.$$ Step 3: Put this into the first equation and calculate $y$: $$-y+4\times 1 = 0\Rightarrow -y+4=0$$ and this gives $$y=4.$$ Step 4: Put this into the second equation and find $x$: $$-x+3\times4-5\times1=7 \Rightarrow -x +7=7$$ and this gives $$x=0.$$ So we can say that the following order of transformations happened: The second row multiplied by $4$ was added to the third row. The places of the first and the second row were exchanged.