## Elementary Linear Algebra 7th Edition

The solution of this system is unique and it is: $$x= 1,\quad y=1,\quad z=-1.$$
As instructed in the problem we will put $A=1/x$, $B=1/y$ and $C=1/z$. This will give us the system of LINEAR equations in $A$, $B$ and $C$: \begin{align*} 2A+B-2C=&5\\ 3A-4B\qquad=&-1\\ 2A+B+3C=&0. \end{align*} Now let us solve this system of equations. Step 1: Subtract the first equation from the third one to eliminate $A$ and $B$: $$2A-2A+B-B+3C-(-2C) = 0-5$$ which becomes $$5C=-5$$ and we get $$C=-1.$$ Step 2: Put this calculated value of $C$ into the other two equations (where it appears) and make the system of two equations in $A$ and $B$: $$2A+B-2(-1) = 5$$ and this becomes $$2A+B+2=5$$ which is finally $$2A+B=3.$$ There was no $C$ in the second equation so we will keep it as it is. Our system now reads: \begin{align*} 2A+B=&3\\ 3A-4B=&-1 \end{align*} Step 3: Express $B$ in terms of $A$ from the first equation: $$B=3-2A.$$ Step 4: Put this into the second equation and solve for $A$: $$3A-4\times(3-2A) = -1\Rightarrow 3A-12+8A = -1\Rightarrow 11A = 11$$ and this gives us $$A=1.$$ Step 5: Use this calculated value to find $B$: $$B= 3-2A = 3-2 = 1.$$ Now let's find $x$, $y$ and $z$. Since $A=1/x$ then $$x=\frac{1}{A} = 1.$$ Since $B=1/y$ then $$y=\frac{1}{B} = 1$$ Since $C=1/z$ then $$z=\frac{1}{C} = \frac{1}{-1} = -1$$