# Solving a system using elementary row operations

• April 24th 2013, 09:56 AM
flametag2
Solving a system using elementary row operations
If someone could please help me with this. its a question about Solving a system using elementary row operations which is honestly one of the easiest and msot simple things to do but for the life of me I cannt understand what I am doing wrong (Headbang)

I solved the system so i know what hte answers are but when doing elementary row operations method i keep getting wrong answers

the Matrix is as per below

3 -2 1 2
1 -1 -1 -4
4 1 -2 0

please if someone could could..just really frustrated at this point :(
• April 24th 2013, 10:25 AM
ebaines
Re: Solving a system using elementary row operations
Show us the first step that you tried and we'll see if you're on the right track.
• April 24th 2013, 12:56 PM
flametag2
Re: Solving a system using elementary row operations
here is how i started the soln

Divide row1 by 3
1 -2/3 1/3 2/3
1 -1 -1 -4
4 1 -2 0

Add (-1 * row1) to row2
1 -2/3 1/3 2/3
0 -1/3 -4/3 -14/3
4 1 -2 0

Add (-4 * row1) to row3
1 -2/3 1/3 2/3
0 -1/3 -4/3 -14/3
0 11/3 -10/3 -8/3

right now im about to divide row 2 by -1/3 and will get back to you soon
• April 24th 2013, 01:18 PM
ebaines
Re: Solving a system using elementary row operations
Looking good so far!
• April 24th 2013, 04:18 PM
Soroban
Re: Solving a system using elementary row operations
Hello, flametag2!

I would avoid fractions when possible.
My first step may surprise you.

Quote:

$\text{Solve the system: }\;\left[\begin{array}{ccc|c} 3&\text{-}2&1&2 \\ 1&\text{-}1&\text{-}1&\text{-}4 \\ 4&1&\text{-}2&0 \end{array}\right]$

$\begin{array}{c}\text{Switch }R_1 \\ \text{and }R_2 \\ \end{array}\left[\begin{array}{ccc|c} 1&\text{-}1&\text{-}1&\text{-}4 \\ 3&\text{-}2&1&2 \\ 4&1&\text{-}2&0 \end{array}\right]$

$\begin{array}{c}\\ R_2-3R_1 \\ R_3-4R_1 \end{array}\left[\begin{array}{ccc|c}1&\text{-}1&\text{-}1&\text{-}4 \\ 0&1&4&14 \\ 0&5&2&16 \end{array}\right]$

$\begin{array}{c}R_1+R_2 \\ \\ R_3-5R_2 \end{array}\left[\begin{array}{ccc|c}1&0&3&10 \\ 0&1&4&14 \\ 0&0&\text{-}18 & \text{-}54 \end{array}\right]$

. . . $\begin{array}{c}\\ \\ \text{-}\frac{1}{18}R_3 \end{array}\left[\begin{array}{ccc|c}1&0&3&10 \\ 0&1&4&14 \\ 0&0&1&3 \end{array}\right]$

$\begin{array}{c}R_1-3R_3 \\ R_2-4R_3 \\ \\ \end{array}\left[\begin{array}{ccc|c}1&0&0&1 \\ 0&1&0&2 \\ 0&0&1&3 \end{array}\right]$

Therefore: . $\begin{Bmatrix}x &=& 1 \\ y &=& 2 \\ z &=&3 \end{Bmatrix}$