Answer
Row Reduced Echelon Form:
\[
\left[ {\begin{array}{cc}
1 & 0 & -5 & 4\\
0& 1 & -6 & 5 \\
\end{array} } \right]
\]
Parametric Form Solutions:
$x_1 = 4+5x_3$
$x_2 = 5+6x_3$
$x_3 $ is free
Work Step by Step
Find the general solutions of the systems
\[
\left[ {\begin{array}{cc}
0 & 1 & -6 & 5\\
1& -2 & 7 & -6 \\
\end{array} } \right]
\]
Swap the first row with the second row
\[
\left[ {\begin{array}{cc}
1 & -2 & 7 & -6\\
0& 1 & -6 & 5 \\
\end{array} } \right]
\]
Pivots are 1 in row 1 and 1 in row 2. Pivot columns in column 1 and 2
Create a zero for the second row pivot by multiplying the second pivot by 2 and adding it to -2
\[
\left[ {\begin{array}{cc}
1 & 0 & -5 & 4\\
0& 1 & -6 & 5 \\
\end{array} } \right]
\]
The matrix is now in reduced echelon form
Define the variables
$x_1 ....-5x_3 = 4$
$..+x_2 -6x_3 = 5$
Solve the first equation for $x_1$ and the second equation for $x_2$. $x_3$ has no restrictions therefore is the free variable.
$x_1 = 4+5x_3$
$x_2 = 5+6x_3$
$x_3 $ is free
