Question 709251
{{{system(2x + y + z = 0,3x - y + z = 3 ,7x - 5y - 3z = 15)}}} 
I do not like the problem either, but no one answered yesterday, so I'll try today. (Warning: I make stupid mistakes in easy calculations, so watch for my mistakes)
Probably you want to use the elimination method, and not matrices.
You want to combine equations to get 2 combination equations with 2 variables.
(We still have a 3-variable-3-equation system, but we can set aside one of the original equations, with all 3 variables, as our third equation, and not worry about it for now).
Then, we would be able to solve the set of combination equations as a two-variable-two-equation system.
 
It looks like the easiest way would be to use the simpler first equation,
combining it with each of the other equations, to eliminate {{{z}}}.
Subtracting the first equation from the second one you get
{{{x-2y=3}}}
Adding {{{3}}} times the first equation to the third one you get
{{{13x-2y=15}}}
Now we solve
{{{system(x-2y=3,13x-2y=15)}}}
I would subtract the new system's first equation from the other equation to get
{{{12x=12}}} --> {{{highlight(x=1)}}}
Substituting into {{{x-2y=3}}} (rather than into {{{13x-2y=15}}} ) looks easy enough.
{{{1-2y=3}}} --> {{{-2y=3-1}}} --> {{{-2y=2}}} --> {{{-2y/(-2)=2/(-2)}}} --> {{{highlight(y=-1)}}}
 
Then, I would go back to one of the original equations and substitute {{{x=1}}} and {{{y=-1}}} to find {{{z}}}.
I chose {{{2x + y + z = 0}}} as the equation that looks easiest.
{{{2(1) + (-1) + z = 0}}} --> {{{2-1+z=0}}} --> {{{1+z=0}}} --> {{{highlight(z=-1)}}}
I verified on my own on scrap paper and it seems to work,
but remember, I warned you that I'm error prone.