I have 2 equations with 3 variables 3x+4y-2z=0 and x-3y+5z=0
if I used elementary transformations \begin{bmatrix}3&4&-2\\1&-3&5\end{bmatrix} I want to express x and y with z so
I substract row 2 - $\frac 13$ row 1 and I get \begin{bmatrix}3&4&-2\\0&-\frac{13}3&-\frac{13}3\end{bmatrix} so in second row x is eliminated and I can express y by z now I want to eliminate y from row 1 so I can express x by z so I add row1 + $\frac3{13}*4 row2$ and I got \begin{bmatrix}3&0&-6\\0&-\frac{13}3&-\frac{13}3\end{bmatrix}
from here 3x=6z => x=2z and $-\frac{13}3y$=$\frac{13}3z$ => y=-z so now if z = k , x=2k and y=-k if I get values of k and replace in equation
for k =1 , x=2 ,y=-1 , z=1 3x+4y-2z=0, 3*2+4*(-1)-2*1 = 0, 6-4-2=0, 0=0 , but in second equation. x-3y+5z=0, 2-3(-1)+5*1=0, 2+3+5=0, 10 is not equal to 0 am I doing something wrong or whats the matter?
multiplying your second equation by $-3$ and adding to the first we get $$x_2=\frac{17}{13}x_3$$ plugtging this in the first one we have $$x_1=-\frac{14}{13}x_3$$ where $x_3$ is an abitrary real number