Question

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 = 2

2x + 4y + 2z = −8

(x, y, z) =

Answer 1

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

$x+2y+z=-4$

$-2x-3y-z=2$

$2x+4y+2z=-8$

Using the Gauss-Jordan elimination method, we get

$\begin{bmatrix}1 & 2 & 1\\ -2 & -3 & -1\\ 2 & 4 & 2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ -8\end{bmatrix}$

$R_3\rightarrow R_3-2R_1$

$\begin{bmatrix}1 & 2 &1\\ -2 & -3 & -1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ 0\end{bmatrix}$

Since elements of bottom row are 0, therefore the system of equations have infinitely many solutions.

$0x+0y+0z=0\Rightarrow 0=0$

$R_2\rightarrow R_2+2R_1$

$\begin{bmatrix}1 & 2 &1\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ -6\\ 0\end{bmatrix}$

$R_1\rightarrow R_1-R_2$

$\begin{bmatrix}1 & 1 &0\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}2\\ -6\\ 0\end{bmatrix}$

$x+y=2$

$y+z=-6$

Let x=t

$t+y=2\rightarrow y=2-t$

The value of y is 2-t.

$(2-t)+z=-6$

$z=-6-2+t$

$z=t-8$

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.