## Elementary Linear Algebra 7th Edition

Apart from the solution $x=0;\quad y=0;\quad z=0$ this system has infinitely many solutions given by $$x= \frac{3}{4}t;\quad y= -2t;\quad z=t.$$
This system always has the solution $x=0;\quad y=0;\quad z=0$ because if you substitute this to each equation all of the terms on the left side will just be equal to zero and that will always exactly match the right side which is also zero. To check if there are other solutions we will transform this system to Row-Echelon form following the steps below: Step 1: Subtract from the second equation the first one: $$12x-12x+4y-5y-z-z = 0$$ which becomes $$-y-2z = 0.$$ Multiply by $-1$ to loose the annoying minuses: $$y+2z = 0.$$ Now our system reads: \begin{align*} 12x+5y+z=&0\\ y+2z=&0 \end{align*} Step 2: Notice that we have only two equations in the Row-Echelon form and three unknowns. This means that we will have infinitely many solutions and to determine them we will equate one of the unknowns, let say $z$ with $t$ that can take any real number as a value and then express the other unknowns in terms of $t$: \begin{align*} 12x+5y+z=&0\\ y+2z=&0\\ z=t \end{align*} Step 3: Substitute $z=t$ to the second equation to get $$y+2t=0$$ and that gives $$y=-2t.$$ Step 4: Substitute $z=t$ and $y=-2t$ to the first equation to get $$12x+5\times(-2t) + t= 0$$ which becomes $$12x-9t=0$$ and that gives $$x= \frac{3}{4}t.$$ Step 5: Notice that beside the solution mentioned in the beginning this system has infinitely many solutions given by $$x= \frac{3}{4}t;\quad y= -2t;\quad z=t.$$