Answer :
Answer:
The system has infinite solutions described in the set [tex]\{(x_1,x_2,x_3)=(6+s1, -10-2s1,s1): s1\in\mathbb{R}\}[/tex]
Step-by-step explanation:
The augmented matrix of the system is [tex]\left[\begin{array}{cccc}-1&-1&-1&4\\2&1&0&2\end{array}\right][/tex].
We apply row operations:
1. We add the first row to the second row twice and obtain the matrix [tex]\left[\begin{array}{cccc}-1&-1&-1&4\\0&-1&-2&10\end{array}\right][/tex]
2. multiply by -1 the rows of the previous matrix and obtain the matrix [tex]\left[\begin{array}{cccc}1&1&1&-4\\0&1&2&-10\end{array}\right][/tex] that is the reduced echelon form of the matrix associated to the system.
Now we aply backward substitution:
1. Observe that the reduced echelon form has a free variable, then the system has infinite solutions.
2.
[tex]x_2+2x_3=-10\\x_2=-10-2x_3[/tex]
3.
[tex]x_1+x_2+x_3=-4\\x_1-10-2x_3+x_3=-4\\x_1-x_3=-4+10\\x_1=6+x_3[/tex].
Then the set of solutions is [tex]\{(x_1,x_2,x_3)=(6+s1, -10-2s1,s1): s1\in\mathbb{R}\}[/tex]