# Thread: MAtrix Problem using Guassian Method

1. ## 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

2. 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.

3. These are linear algebra questions! Stop posting these in Advanced Applied Math!

4. 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.

5. 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.

6. 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

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

7. 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?

8. 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

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

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

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

$b=\frac{13}{5}-\frac{3}{5}$
$b=2$

9. yes it's correct 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

edit: I apologize, I'm not gonna do that again

10. 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.)