0

$$5x-y+3z=3$$$$-3x+y-z=1$$$$-4x+3y+2z=9$$ <=> I take $$3*(-y)+y=-2y$$ and $$3*3z+(-z)=8z$$ and $$3*3+1=10$$third $$4*(-y)+3y=-y$$ and $$4*3z+2z=14z$$ and $$4*3+9=21$$ which results in $$5x-y+3z=3$$$$-2y+8z=10$$$$-y+14z=21$$ and THAT takes away the x from equation 1 and 2. Further on same equation1 and 2 $$5x-y+3z=3$$$$-2y+8y=10$$ and $$1*8-2*14=-20$$$$1*10-2*21=-32$$ and that is equation 3 $$5x-y+3z=3$$$$-2y+8y=10$$$$-20z=-32$$ and that means that z is $$\frac{-32}{-20}$$ shortened $$z=\frac{16}{10}$$ is this correct so far? I know that i can get x and y as well but im asking if this is correct and if so I can get x and y later.

user1838781
  • 457
  • $z=1$ and not $\dfrac{1}{3}$. Solve it again and you might get the correct answers this time. – lsp Apr 23 '13 at 11:11
  • 2
    ...and next time you post a question and more important: that you write an exam, it'll be a good idea if you state clearly what you did in each step. I can't talk for all universities or math departments or lecturers, but for me it always was pretty simple: anything I can't understand because of the student's lack of desire to write clearly is going to be sadly and minimally graded. – DonAntonio Apr 23 '13 at 11:16

2 Answers2

2

Since I don't like fractions, I would add the 2nd equation to the 1st equation, getting $$2x+2z=4\tag4$$ and subtract $3$ times the 2nd equation from the 3rd, getting $$5x+5z=6\tag5$$ Now (4) says $x+z=2$, while (5) says $x+z=6/5$, and they can't both be right, so there is no solution.

Gerry Myerson
  • 179,216
1

It looks like your mistake is in the first step. Here is the solution: $$5x−y+3z=3$$ $$−3x+y−z=1$$ $$−4x+3y+2z=9$$ $3/5*R_1 + R_2 \rightarrow R_2$
$4/5*R_1 + R_3 \rightarrow R_3$ $$5x-y+3z=3$$ $$2/5* y+4/5*z=14/5$$ $$11/5*y+22/5*z=57/5$$ $-11/2*R_2+R_3 \rightarrow R_3$ $$2/5* y+4/5*z=14/5$$ $$11/5*y+22/5*z=57/5$$ $$0=4$$

Contradiction, no solution.
Edited to change to fractions.

user2307487
  • 178
  • Why do you divide 3 and 5 and 11? – user1838781 Apr 23 '13 at 12:51
  • Multiplying the 1st equation by $3/5$ and adding the result to the 2nd equation gets rid of the $x$ term in the 2nd equation. Similar explanations for the other actions. – Gerry Myerson Apr 23 '13 at 13:06
  • I would prefer to see the fractions left as fractions instead of decimals. In this case, with all the denominators 2 and 5, the decimals are exact. If the denominators were other things, they are not, while the fractions are. But the approach is good and the answer correct. – Ross Millikan Apr 23 '13 at 13:18
  • @user1838781 : Yes, exactly what Gerry said. You should always label your matrix operations similar to the way I did so others can follow yours steps. Also your edited solution to the problem is still wrong as of the time of this comment. Please add the steps you are using and I will point out any errors. – user2307487 Apr 23 '13 at 13:21
  • @user2307487 Can you write your answer in fractions instead of decimals, please its so much easier to follow your solution then – user1838781 Apr 23 '13 at 13:43
  • 1
    @user1838781 : I can see what's happening now. The steps you are doing are correct ONLY when equation (1) has an x coefficient of 1. With the x coefficient being 5, it does not cancel in equations (2) and (3) when you multiply by 3 and 4 respectively. This is the reason I used fractions. – user2307487 Apr 23 '13 at 13:52