2.3: Reduced Row Echelon Form

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If we continue the row elimination procedure so that all the pivots are one, and all the entries above and below the pivot are eliminated, then the resulting matrix is in the so-called reduced row echelon form. We write the reduced row echelon form of a matrix \(\text{A}\) as \(\text{rref}(\text{A})\). If \(\text{A}\) is an invertible square matrix, then \(\text{rref}(\text{A}) = \text{I}\).

Instead of Gaussian elimination and back substitution, a system of equations can be solved by bringing a matrix to reduced row echelon form. We can illustrate this by solving again our first example. Beginning with the same augmented matrix, we have

\[\begin{array}{l}\left(\begin{array}{rrrr}-3&2&-1&-1\\6&-6&7&-7\\3&-4&4&-6\end{array}\right)\to\left(\begin{array}{rrrr}-3&2&-1&-1\\0&-2&5&-9\\0&-2&3&-7\end{array}\right)\to\left(\begin{array}{rrrr}-3&0&4&-10\\0&-2&5&-9\\0&0&-2&2\end{array}\right) \\ \to\left(\begin{array}{rrrr}-3&0&4&-10\\0&-2&5&-9\\0&0&1&-1\end{array}\right)\to\left(\begin{array}{rrrr}-3&0&0&-6\\0&-2&0&-4\\0&0&1&-1\end{array}\right)\to\left(\begin{array}{rrrr}1&0&0&2\\0&1&0&2\\0&0&1&-1\end{array}\right).\end{array}\nonumber \]

Once \(\text{A}\) has been transformed into the identity matrix, the resulting system of equations is just the solution, that is, \(x_1 = 2,\: x_2 = 2\) and \(x_3 = −1\).