2. Use Gauss-Jordan Elimination to find the solution to the system or show that none exists.
$$\begin{array}{rl}x-3y+z& =1\\ -2x+y+3z& =-7\\ x-4y+2z& =0\end{array}$$
a. Solve the system using the Gauss-Jordan Elimination Method. Show all row operations and the final matrix in reduced row echelon form.
b. Write the solution as an ordered triple. $x=-1,y=2,7=3$
c. Fill out the table to show two of the possible solutions to the system.
Hint: Pick two random values for $z$ to generate the ordered triples.

Step 1 ：Write the system of equations as an augmented matrix: $$\left[\begin{array}{cccccccccc}1& -3& 1& 1-2& 1& 3& -71& -4& 2& 0\end{array}\right]$$

Step 2 ：Swap row 1 and row 2 to get the largest absolute value in the top left corner: $$\left[\begin{array}{cccccccccc}-2& 1& 3& -71& -3& 1& 11& -4& 2& 0\end{array}\right]$$

Step 3 ：Multiply row 1 by -1/2 to get a 1 in the top left corner: $$\left[\begin{array}{cccccccccc}1& -0.5& -1.5& 3.51& -3& 1& 11& -4& 2& 0\end{array}\right]$$

Step 4 ：Subtract row 1 from row 2 and row 3 to get zeros below the leading 1 in the first column: $$\left[\begin{array}{cccccccccc}1& -0.5& -1.5& 3.50& -2.5& 2.5& -2.50& -3.5& 3.5& -3.5\end{array}\right]$$

Step 5 ：Multiply row 2 by -1/2.5 and row 3 by -1/3.5 to get a 1 in the second and third row of the second column: $$\left[\begin{array}{cccccccccc}1& -0.5& -1.5& 3.50& 1& -1& 10& 1& -1& 1\end{array}\right]$$

Step 6 ：Subtract row 2 from row 3 to get a zero in the third row of the second column: $$\left[\begin{array}{cccccccccc}1& -0.5& -1.5& 3.50& 1& -1& 10& 0& 0& 0\end{array}\right]$$

Step 7 ：Write the system of equations from the final matrix: $$\begin{array}{rlrl}x-0.5y-1.5z& =3.5y-z& =10& =0\end{array}$$

Step 8 ：Solve for x, y, and z to get the solution as an ordered triple: $x,y,z$ = $3.5+0.5y+1.5z,1+z,z$

Step 9 ：Choose two random values for z to find two possible solutions to the system. For example, if z = 0, then $x,y,z$ = $3.5+0.5(1)+1.5(0),1+0,0$ = $4,1,0$. If z = 1, then $x,y,z$ = $3.5+0.5(1+1)+1.5(1),1+1,1$ = $6,2,1$

Step 10 ：So, two possible solutions to the system are $\boxed{{\displaystyle (4,1,0)}}$ and $\boxed{{\displaystyle (6,2,1)}}$