Chapter 9 - Systems and Matrices - Summary Exercises on Systems of Equations - Exercises - Page 907: 15

$\{(0,1,0)\}$

1. Add the second and third equations to get $2y-3z=2$ 2. Reduce the second equation (divide by 2) to get $-2x+y-3z=1$ and add it to the first equation to get $4y+z=4$ 3. Multiply 3 to the above to get $12y+3z=12$ and add it to the equation from step-1 to get $14y=14$, thus $y=1$ 4. We can get $z=4-4y=0$ and $4x=-3z=0$, thus $x=0$ 5. Thus the solution set is $\{(0,1,0)\}$