Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?
$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$
Every time I solve this it feels like it could have been done simpler... Thanks!
I first exchanged row 1 and row 2 to get:
$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$ $$3x_1-2x_2+3x_3=8$$
Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:
$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$
But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!