## 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 $$x=-\frac{3}{5}t;\quad y= \frac{4}{5}t;\quad z=t.$$ where $t$ can have any real numbered value.
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 find out if there are other solutions we will transform this system to Row-Echelon form following the steps below: Step 1: From third equation subtract the first one $$5x-5x + 15y-5y-9z-(-z)=0$$ which becomes $$10y-8z=0.$$ we can divide by $2$ to simplify: $$5y-4z = 0.$$ Now our system reads \begin{align*} 5x+5y-z = &0\\ 10x+5y+2z = &0\\ 5y-4z=&0. \end{align*} Step 2: From 2nd equation subtract the first one multiplied by $2$: $$10x-2\times5x + 5y-2\times 5y +2z-2\times (-z) = 0$$ which becomes $$-5y+4z=0.$$ Our system now reads \begin{align*} 5x+5y-z=&0\\ -5y+4z=&0\\ 5y-4z=&0 \end{align*} Step 3: Notice that the last equation is just the second one multiplied by $-1$ so it does not give any new conditions and we can drop it. Now we have two equations and three unknowns which means that we will have infinitely many solutions. To find them we will equate one of the unknowns, lets 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*} 5x+5y-z=&0\\ -5y+4z=&0\\ z=t \end{align*} Step 4: Substitute $z= t$ to the second equation and express $y$ in terms of $t$: $$-5y+4t=0$$ and that gives $$y= \frac{4}{5}t.$$ Step 5: Substitute $z=t$ and $y=\frac{4}{5}t$ to the first equation and express $x$ in terms of $t$: $$5x+5\times\frac{4}{5}t-t = 0$$ and that gives $$5x+3t=0$$ which is finally $$x=-\frac{3}{5}t.$$ Step 5: We see that appart from the solution mentioned at the beginning there are infinitely many solutions given by $$x=-\frac{3}{5}t;\quad y= \frac{4}{5}t;\quad z=t.$$