Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answer in terms of the parameters t and/or s.)x + 2y + z = −4−2x − 3y − z = 22x + 4y + 2z = −8(x, y, z) =

Answer:The system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.Step-by-step explanation:The given educations are[tex]x+2y+z=-4[/tex][tex]-2x-3y-z=2[/tex][tex]2x+4y+2z=-8[/tex]Using the Gauss-Jordan elimination method, we get[tex]\begin{bmatrix}1 & 2 & 1\\ -2 & -3 & -1\\ 2 & 4 & 2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ -8\end{bmatrix}[/tex][tex]R_3\rightarrow R_3-2R_1[/tex][tex]\begin{bmatrix}1 & 2 &1\\ -2 & -3 & -1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ 0\end{bmatrix}[/tex]Since elements of bottom row are 0, therefore the system of equations have infinitely many solutions.[tex]0x+0y+0z=0\Rightarrow 0=0[/tex][tex]R_2\rightarrow R_2+2R_1[/tex][tex]\begin{bmatrix}1 & 2 &1\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ -6\\ 0\end{bmatrix}[/tex][tex]R_1\rightarrow R_1-R_2[/tex][tex]\begin{bmatrix}1 & 1 &0\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}2\\ -6\\ 0\end{bmatrix}[/tex][tex]x+y=2[/tex][tex]y+z=-6[/tex]Let x=t[tex]t+y=2\rightarrow y=2-t[/tex]The value of y is 2-t.[tex](2-t)+z=-6[/tex][tex]z=-6-2+t[/tex][tex]z=t-8[/tex]The value of z is t-8.Therefore the he system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.