## Elementary Linear Algebra 7th Edition

The difference between the reduced row-echelon form and reduced echelon form is that; by using the reduced row-echelon form one can get the solution of the corresponding system directly. But by using the reduced echelon form, one has to use the back substitution to get the solution of the corresponding system. For example, the following augmented matrix in the reduced row-echelon form $$\left[ \begin{array} {cc} 1&0&1\\0&1&-2 \end{array} \right].$$ The corresponding system is \begin{align*} x&=1\\ y&=-2.\\ \end{align*} Which gives the solution $x=1, y=-2$. But, the following augmented matrix in the row-echelon form $$\left[ \begin{array} {cc} 1&2&1\\0&3&3 \end{array} \right].$$ The corresponding system is \begin{align*} x+2y&=1\\ 3y&=3.\\ \end{align*} Using the back substitution we obtain the solution $x=-1, y=1$.