Use Gauss-Jordan to solve 4-variable linear system

[mooshupork34]

The problem is: Use the Gauss-Jordan method to find the complete solution set for the following system, and express the solution set as a linear combination.

2x1 - 6x2 + 3x3 - 21x4 = 0
4x1 - 5x2 + 2x3 - 24x4 = 0
-x1 + 3x2 - x3 + 10x4 = 0
-2x1 + 3x2 - x3 + 13x4 = 0

So I transformed this system into a matrix.

2 - 6 3 - 21 | 0
4 - 5 2 - 24 | 0
-1 3 -1 10 | 0
-2 3 -1 13 | 0

I'm not really sure where to go fom here. I know that you are supposed to use Row Reduced Echelon Form, but every time I do it, I keep getting an answer other than the one the calculator gives me, which leads me to think that I'm doing it wrong. If anyone could show me how to approach this, I would greatly appreciate it.

 

[mark07]

\(\displaystyle \L \left[
\begin{matrix}
2 & -6 & 3 & -21 & 0 \\
4 & -5 & 2 & -24 & 0 \\
-1 & 3 & -1 & 10 & 0 \\
-2 & 3 & -1 & 13 & 0
\end{matrix}
\right]\)

Interchange rows 1 and 3 and then multiply row 1 by (-1):

\(\displaystyle \L \left[
\begin{matrix}
1 & -3 & 1 & -10 & 0 \\
4 & -5 & 2 & -24 & 0 \\
2 & -6 & 3 & -21 & 0 \\
-2 & 3 & -1 & 13 & 0
\end{matrix}
\right]\)


add -4 times row 1 to row 2,
add -2 times row 1 to row 3,
add 2 times row 1 to row 4,

\(\displaystyle \L \left[
\begin{matrix}
1 & -3 & 1 & -10 & 0 \\
0 & 7 & -2 & 16 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & -3 & 1 & -7 & 0
\end{matrix}
\right]\)



add 2 times row 4 to row 2 (instead of dividing row 2 by 7 and dealing with fractions)

\(\displaystyle \L \left[
\begin{matrix}
1 & -3 & 1 & -10 & 0 \\
0 & 1 & 0 & 2 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & -3 & 1 & -7 & 0
\end{matrix}
\right]\)



add 3 times row 2 to row 4

\(\displaystyle \L \left[
\begin{matrix}
1 & -3 & 1 & -10 & 0 \\
0 & 1 & 0 & 12 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 & 0
\end{matrix}
\right]\)


add -1 times row 3 to row 4

\(\displaystyle \L \left[
\begin{matrix}
1 & -3 & 1 & -10 & 0 \\
0 & 1 & 0 & 12 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{matrix}
\right]\)

The zero row tells you that there are infinitely many solutions. Let x_4 be free (since it corresponds to a non-pivot column), i.e. \(\displaystyle x_4=t\), then

\(\displaystyle \L x_3 = t\)
\(\displaystyle \L x_2 =-12t\)
\(\displaystyle \L x_1 = -27t\)

and the solution space is of the form

\(\displaystyle \L \left[
\begin{matrix}
-27 \\
-12 \\
1 \\
1
\end{matrix}
\right] \; t\)

for real t.