#math/linear-algebra Row reduction replaces a system with simpler equations without changing its solution set. Consider $ \begin{cases} x_1+2x_2+3x_3=1,\\ 3x_1+2x_2+x_3=7,\\ 2x_1+x_2+2x_3=1. \end{cases} $ The forward phase clears entries below the pivots. Subtract multiples of the first row, scale the second row, and use it to clear the entry below its pivot: $ \left(\begin{array}{ccc|c} 1&2&3&1\\ 3&2&1&7\\ 2&1&2&1 \end{array}\right) \sim \left(\begin{array}{ccc|c} 1&2&3&1\\ 0&1&2&-1\\ 0&0&2&-4 \end{array}\right). $ Here $\sim$ means the matrices are row equivalent. The resulting `echelon form` lets us solve from bottom to top by back substitution: $ \begin{aligned} x_3&=-2,\\ x_2&=-1-2x_3=3,\\ x_1&=1-2x_2-3x_3=1. \end{aligned} $ Alternatively, continue with the backward phase: scale the last row to make its pivot $1$, then clear the entries above each pivot, working upward. This gives `reduced echelon form`: $ \left(\begin{array}{ccc|c} 1&0&0&1\\ 0&1&0&3\\ 0&0&1&-2 \end{array}\right). $ Now we can read the solution directly: $\mathbf{x}=(1,3,-2)^T$.