Solving a System of Equations By Reduction to Row Reduced Echelon Form
Suppose we have a system of equations\[2x_1+2x_2-x_3+x_5=0\]
\[-x_1-x_2-2x_3-3x_4+x_5=0\]
\[x_1+x_2-2x_3-x_5=0\]
\[x_3+x_4+x_5=0\]
We can represent this in matrix form as
\[ \left( \begin{array}{ccccc} 2 & 2 & -1 & 0 & 1 \\ -1 & -1 & 2 & -3 & 1 \\ 1 & 1 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \end{array} \right) \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = \begin{pmatrix}0\\0\\0\\0\\0\end{pmatrix} \]
Now row reduce the matrix
\[ \left( \begin{array}{ccccc} 2 & 2 & -1 & 0 & 1 \\ -1 & -1 & 2 & -3 & 1 \\ 1 & 1 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \end{array} \right)\]
Divide the first row by 2.
\[ \left( \begin{array}{ccccc} 1 & 1 & -1/2 & 0 & 1/2 \\ -1 & -1 & 2 & -3 & 1 \\ 1 & 1 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \end{array} \right)\]
Add row 1 to row 2 and subtract row 1 from row 3.
\[ \left( \begin{array}{ccccc} 1 & 1 & -1/2 & 0 & 1/2 \\ 0 & 0 & 3/2 & -3 & 3/2 \\ 0 & 0 & -3/2 & 0 & -3/2 \\ 0 & 0 & 1 & 1 & 1 \end{array} \right)\]
Divide row 2 by 3/2.
\[ \left( \begin{array}{ccccc} 1 & 1 & -1/2 & 0 & 1/2 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & -3/2 & 0 & -3/2 \\ 0 & 0 & 1 & 1 & 1 \end{array} \right)\]
Add half of row 2 to row 1, add three halves of row 2 to row 3, and subtract row 2 from row 4
\[ \left( \begin{array}{ccccc} 1 & 1 & 0 & -1 & 1 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & 3 & 0 \end{array} \right)\]
Divide row 3 by -3
\[ \left( \begin{array}{ccccc} 1 & 1 & 0 & -1 & 1 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 & 0 \end{array} \right)\]
Add row 3 to row 1, add twice row 3 to row 2 and subtract three times row 3 from row 4.
\[ \left( \begin{array}{ccccc} 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right)\]
This matrix is in row reduced echelon form. We obtain
\[x_1+x_2+x_5=0\]
\[x_3+x_5=0\]
\[x_4=0\]
We can set \[x_2=s, \; x_5=t\]
then \[x_1=-s-t, \; x_2 =s, \: x_3=-t, \; x_4 =0, \; x_5=t\]
where \[s, \; t\]
are arbitrary.