2. Use Gauss-Jordan Elimination to find the solution to the system or show that none exists. \[ \begin{aligned} x-3 y+z & =1 \\ -2 x+y+3 z & =-7 \\ x-4 y+2 z & =0 \end{aligned} \] 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: \[\begin{bmatrix} 1 & -3 & 1 & 1 \ -2 & 1 & 3 & -7 \ 1 & -4 & 2 & 0 \end{bmatrix}\]

Step 2 ：Swap row 1 and row 2 to get the largest absolute value in the top left corner: \[\begin{bmatrix} -2 & 1 & 3 & -7 \ 1 & -3 & 1 & 1 \ 1 & -4 & 2 & 0 \end{bmatrix}\]

Step 3 ：Multiply row 1 by -1/2 to get a 1 in the top left corner: \[\begin{bmatrix} 1 & -0.5 & -1.5 & 3.5 \ 1 & -3 & 1 & 1 \ 1 & -4 & 2 & 0 \end{bmatrix}\]

Step 4 ：Subtract row 1 from row 2 and row 3 to get zeros below the leading 1 in the first column: \[\begin{bmatrix} 1 & -0.5 & -1.5 & 3.5 \ 0 & -2.5 & 2.5 & -2.5 \ 0 & -3.5 & 3.5 & -3.5 \end{bmatrix}\]

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: \[\begin{bmatrix} 1 & -0.5 & -1.5 & 3.5 \ 0 & 1 & -1 & 1 \ 0 & 1 & -1 & 1 \end{bmatrix}\]

Step 6 ：Subtract row 2 from row 3 to get a zero in the third row of the second column: \[\begin{bmatrix} 1 & -0.5 & -1.5 & 3.5 \ 0 & 1 & -1 & 1 \ 0 & 0 & 0 & 0 \end{bmatrix}\]

Step 7 ：Write the system of equations from the final matrix: \[\begin{aligned} x - 0.5y - 1.5z & =3.5 \ y - z & =1 \ 0 & =0 \end{aligned}\]

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 \(\boxed{(4, 1, 0)}\) and \(\boxed{(6, 2, 1)}\)