# MAtrix Problem using Guassian Method

• July 23rd 2010, 02:57 AM
Paymemoney
MAtrix Problem using Guassian Method
Hi

Can someone tell me what should be the next step:
I have tried many ways but cannot seem to get the right answer.
This is what i have done:

$\begin{bmatrix}
1 & 2 & 1 & | & 6 \\
2 & -1 & -1 & |& -1\\
3 & 2 & 1 & | & 8
\end{bmatrix}$

2R1-R2 -> R2
3R1-R3 -> R3

$\begin{bmatrix}
1 & 2 & 1 & | & 6 \\
0 & 6 & 3 & |& 13 \\
0 & 4 & 2 & | & 10
\end{bmatrix}$

P.S
• July 23rd 2010, 03:03 AM
Ackbeet
Well, on a computer, you wouldn't multiply row 2 by 1/6, but when doing things by hand, that's what I would do next. It makes it slightly easier to contemplate the next step of getting a zero in the 3,2 position.
• July 23rd 2010, 03:03 AM
Ackbeet
These are linear algebra questions! Stop posting these in Advanced Applied Math!
• July 23rd 2010, 04:28 AM
Paymemoney
Quote:

Originally Posted by Ackbeet
These are linear algebra questions! Stop posting these in Advanced Applied Math!

But shouldn't this be posted here, i was told by mod, that on one of my previous post on matrix it was in pre-algebra section, however in fact it should of been in Applied Math.
• July 23rd 2010, 04:37 AM
Ackbeet
The pre-algebra forum is in the Pre-University Math Help forum. I guarantee you that your problems would fit correctly in the University Math Help/Linear and Abstract Algebra subforum.
• July 23rd 2010, 05:52 PM
Paymemoney
After multiplying by 1/6 to row 2, i then done the following, however i get stuck again because i end up with:

$a+2b+c = 6$
$b+\frac{1}{2}c = \frac{13}{6}$
$a=1$

Do you have any suggestions of what i should do instead?

$\begin{bmatrix}
1 & 2 & 1 & | & 6 \\
0 & 1 & \frac{1}{2} & | & \frac{13}{6} \\
0 & 4 & 2 & | & 10
\end{bmatrix}$

R3-2R1 -> R3

$\begin{bmatrix}
1 & 2 & 1 & | & 6 \\
0 & 1 & \frac{1}{2} & | & \frac{13}{6} \\
-2 & 0 & 0 & | & -2
\end{bmatrix}$
• July 24th 2010, 02:16 AM
Ackbeet
Hmm. I double-checked your earlier work, and I think you made a mistake. Your original system is this:

$
\left[\begin{matrix}1&2&1\\2&-1&-1\\3&2&1\end{matrix}\left|\begin{matrix}6\\-1\\8\end{matrix}\right].
$

After performing the following elementary row operations, as you suggested:

2R1-R2 -> R2
3R1-R3 -> R3,

I get this:

$
\left[\begin{matrix}1&2&1\\0&5&3\\0&4&2\end{matrix}\left |\begin{matrix}6\\13\\10\end{matrix}\right].
$

The reason I thought there was an error, was because the original matrix definitely has a solution, but your simplified matrix after those two row operations was inconsistent. So, at this point, I would divide row 2 by 5. What do you get then?
• July 24th 2010, 05:12 AM
Paymemoney
This is want i got when i divided by 5:

$\begin{bmatrix}
1 & 2 & 1 & | & 6\\
0 & 1 & \frac{3}{5} & | & \frac{13}{5}\\
0 & 4 & 2 & | & 10
\end{bmatrix}$

R3-4R2 -> R3

$\begin{bmatrix}
1 & 2 & 1 & | & 6\\
0 & 1 & \frac{3}{5} & | & \frac{13}{5}\\
0 & 0 & \frac{-2}{5} & | & \frac{-2}{5}
\end{bmatrix}$

$a+2b+c=6$
$a=1$

$b+\frac{3}{5}$
$c=\frac{13}{5}$
$c=\frac{-2}{5}*\frac{-5}{21}$
$c=1$

$b=\frac{13}{5}-\frac{3}{5}$
$b=2$
• July 24th 2010, 05:20 AM
yeKciM
yes it's correct :D u always when u get some results just put them in your starting system of equations and see is it solution that u get the right one :D

edit: I apologize, I'm not gonna do that again :D
• July 24th 2010, 07:20 AM
Ackbeet
Concur with yeKciM. Incidentally, yeKciM: on this forum it's considered bad form to use anything but full and correct English. Please don't use chat-lingo or texting abbreviations! Thanks! (For reference, look at Rule #12 here.)