## Elementary Linear Algebra 7th Edition

This system has unique solution and it is given by $$x=1;\quad y= 0;\quad z= 3;\quad w=2.$$
To transform this system to Row-Echelon form we will follow the steps below Step 1: Eliminate $x$ from the second equation by subtracting from it the first equation multiplied by $2$: $$2x-2x + 3y-2y+0z-2z -w-2w = 0-12$$ which is then $$y-2z-3w = -12.$$ Step 2: Eliminate $x$ from the 3rd equation by adding to it the first one multiplied by $3$: $$-3x+3x+4y+3y+z+3z+2w+3w = 4+3\times6$$ which becomes $$7y+4z+5w = 22$$ Step 3: Eliminate $x$ from the last equation by subtracting the first one from it: $$x-x+2y-y-z-z+w-w = 0-6$$ which becomes $$y-2z=-6.$$ Step 4: Now after we have done our eliminations we will just rewrite what we have: \begin{align*} x+y+z+w = &6\\ y-2z-3w=&-12\\ 7y+4z+5w=&22\\ y-2z\qquad=&-6 \end{align*} Step 5: Observe that we can directly get $w$ by subtracting the 2nd equation from the 4th: $$y-y-2z-(-2z)+0w-(-3w) = -6-(-12)$$ which becomes $$3w = 6$$ and that is finally $$w=2.$$ Step 6: Put in the solution for $w$ in each equation and simplify it: 1st: $x+y+z+2=6\Rightarrow x+y+z = 4$ 2nd: $y-2z-3\times 2 = -12 \Rightarrow y-2z= -6$ 3rd: $7y+4z+5\times2 = 22\Rightarrow 7y+4z=12$ 4th: $y-2z= -6$ Step 7: Rewrite the system by dropping the 4th equation since it is identical to the 2nd one: \begin{align*} x+y+z=&4\\ y-2z=&-6\\ 7y+4z=&12 \end{align*} Step 8: To eliminate $y$ from the 3rd equation substitute from it the second one multiplied by $7$: $$7y-7y+4z-7\times(-2z)= 12-7\times(-6)$$ which becomes $$18z= 54$$ and that gives $$z = 3.$$ Step 9: Substitute the calculated value for $z$ into the second equation and find $y$: $$y-2\times3 = -6$$ which becomes $$y = 0.$$ Step 10: Substitute calculated values for $y$ and $z$ into the 1st equation and find $x$: $$x+0+3=4$$ which gives $$x=1.$$ Step 11: Summarize all of the obtained results and write down the solution of the system: $$x=1;\quad y= 0;\quad z= 3;\quad w=2.$$