As we saw in The Matrix and Solving Systems using Matrices section, the reduced row echelon form method can be used to solve systems.
With this method, we put the coefficients and constants in one matrix (called an augmented matrix, or in coefficient form) and then, with a series of row operations, change it into what we call reduced echelon form, or reduced row echelon form. This is when the leading entry in every nonzero row is 1, and each leading 1 is the only nonzero entry for that column. This is also called Gaussian Elimination, or Row Reduction.
To get the matrix in the correct form, we can 1) swap rows, 2) multiply rows by a nonzero constant, or 3) replace a row with the product of another row times a constant added to the row to be replaced.
We can do this “by hand” by doing the following, along with an example of solving the system (sorry about all the fractions!):
\(\displaystyle \begin{array}{l}5x6y\,\,7z\,\,\,\,=\,\,\,\,7\\6x4y+10z=\,34\\2x+4y\,\,\,\,3z\,\,=\,\,\,29\end{array}\)
 Find the first nonzero entry in the first column; this is the pivot. Move this to the first row.
This is just 5, so the augmented matrix is still \(\displaystyle \left[ {\begin{array}{*{20}{c}} 5 & {6} & {7} & 7 \\ 6 & {4} & {10} & {34} \\ 2 & 4 & {3} & {29} \end{array}} \right]\,\).
 Make the pivot (the 5) 1 by multiplying this row by \(\displaystyle \frac{1}{5}\): \(\displaystyle \left[ {\begin{array}{*{20}{c}} 5 & {6} & {7} & 7 \\ 6 & {4} & {10} & {34} \\ 2 & 4 & {3} & {29} \end{array}} \right]\,\begin{array}{*{20}{c}} {\frac{1}{5}R1\Rightarrow R1} \\ {} \end{array}\).
Now we have \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 6 & {4} & {10} & {34} \\ 2 & 4 & {3} & {29} \end{array}} \right]\).
 Now we want the elements below the 1 (first column) to be 0. To get them to be 0, add multiplies of the first (pivot) row to each of the rows below, replacing those rows:
\(\displaystyle \begin{array}{c}\left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 6 & {4} & {10} & {34} \\ 2 & 4 & {3} & {29} \end{array}} \right]\begin{array}{*{20}{c}} {} \\ {6R1\text{ }+\text{ }R2\text{ }\Rightarrow \text{ }R2} \\ {2R1\text{ }+\text{ }R3\text{ }\Rightarrow \text{ }R3} \end{array}\,\,\,\,\\=\left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ {6\cdot 1+6=0} & {6\cdot \frac{6}{5}4=\frac{{16}}{5}} & {6\cdot \frac{7}{5}+10=\frac{{92}}{5}} & {6\cdot \frac{7}{5}34=\frac{{212}}{5}} \\ {2\cdot 1+2=0} & {2\cdot \frac{6}{5}+4=\frac{{32}}{5}} & {2\cdot \frac{7}{5}3=\frac{1}{5}} & {2\cdot \frac{7}{5}+29=\frac{{131}}{5}} \end{array}} \right]\end{array}\)
Now we have \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 0 & {\frac{{16}}{5}} & {\frac{{92}}{5}} & {\frac{{212}}{5}} \\ 0 & {\frac{{32}}{5}} & {\frac{1}{5}} & {\frac{{131}}{5}} \end{array}} \right]\).

 Now the second row will be the pivot row. To get the second column, second row a 1, divide this row by \(\frac{{16}}{5}\) (multiply by reciprocal): \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 0 & {\frac{{16}}{5}} & {\frac{{92}}{5}} & {\frac{{212}}{5}} \\ 0 & {\frac{{32}}{5}} & {\frac{1}{5}} & {\frac{{131}}{5}} \end{array}} \right]\begin{array}{*{20}{c}} {} \\ {\frac{5}{{16}}R2} \\ {} \end{array}\Rightarrow R2\).
Now we have \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & {\frac{{32}}{5}} & {\frac{1}{5}} & {\frac{{131}}{5}} \end{array}} \right]\,\,\).
 Repeat the procedure from Step 3 above, using the first and third rows, to make their second column entries 0: \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & {\frac{6}{5}} & {\frac{7}{5}} & {\frac{7}{5}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & {\frac{{32}}{5}} & {\frac{1}{5}} & {\frac{{131}}{5}} \end{array}} \right]\begin{array}{*{20}{c}} {\frac{6}{5}R2+R1\Rightarrow R1} \\ {} \\ {\frac{{32}}{5}R2+R3\Rightarrow R3} \end{array}\).
Now we have \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & 0 & {\frac{{11}}{2}} & {\frac{{29}}{2}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & 0 & {37} & {111} \end{array}} \right]\,\).
 Now multiply the third row by \(\frac{1}{{37}}\) to have the 1 in the third column for another pivot row:
\(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & 0 & {\frac{{11}}{2}} & {\frac{{29}}{2}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & 0 & {37} & {111} \end{array}} \right]\,\,\,\begin{array}{*{20}{c}} {} \\ {} \\ {\frac{1}{{37}}R3\Rightarrow R3} \end{array}\).
Now we have \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & 0 & {\frac{{11}}{2}} & {\frac{{29}}{2}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & 0 & 1 & {3} \end{array}} \right]\), which is in row echelon form.
 Repeat step 3 above again to get 0’s in the third column (first two rows) to get reduced row echelon form: \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & 0 & {\frac{{11}}{2}} & {\frac{{29}}{2}} \\ 0 & 1 & {\frac{{23}}{4}} & {\frac{{53}}{4}} \\ 0 & 0 & 1 & {3} \end{array}} \right]\begin{array}{*{20}{c}} {\frac{{11}}{2}R3+R1\Rightarrow R1} \\ {\frac{{23}}{4}R3+R2\Rightarrow R2} \\ {} \end{array}\,\).
We get \(\displaystyle \left[ {\begin{array}{*{20}{c}} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & {3} \end{array}} \right]\).
Yeah! Our answer is \(x=2, y=4\), and \(z=3\)!
Note if at the end we get all 0’s in the last row, we have an infinite number of solutions, and if we get all 0’s except for the last row, fourth column, we have no solution. We saw these examples here in the Matrix and Solving Systems with Matrices section.