SOLUTION: Solve using matrices. 2x-y-z=4, x+y-5z=-4, x-2y=4 I'm trying not to use fractions like the solution key.I tried using the following: R3-R2=R3, R1-2R2=R2, 1/3R2, 3R2+R3=R3, 14R1+R3

Algebra ->  Matrices-and-determiminant -> SOLUTION: Solve using matrices. 2x-y-z=4, x+y-5z=-4, x-2y=4 I'm trying not to use fractions like the solution key.I tried using the following: R3-R2=R3, R1-2R2=R2, 1/3R2, 3R2+R3=R3, 14R1+R3      Log On


   

Question 1074117: Solve using matrices. 2x-y-z=4, x+y-5z=-4, x-2y=4 I'm trying not to use fractions like the solution key.I tried using the following: R3-R2=R3, R1-2R2=R2, 1/3R2, 3R2+R3=R3, 14R1+R3=R1, 14R2-3R3=R2, R1+R2=R1 AND THEN I get a wrong solution. What am i doing wrong?
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
The steps being abbreviated here:

2E2-E1 and E3-E2 to change E2 and E3

E2 divided by 3

E3+3E2 to change E2

-1*E3

E1-E3 and E2+3E3 to change E1 and E2

E1+E2 to change E1

Resulting matrix became
%28matrix%283%2C4%2C%0D%0A2%2C0%2C0%2C5%2C%0D%0A0%2C1%2C0%2C-1%2C%0D%0A0%2C0%2C1%2C1%29%29

This seems that you cannot completely avoid fractions, none of my steps up to this last part used fractions. You will need a fraction to do the last step to finish.