# Quick Lin Alg question

• Sep 4th 2008, 09:22 PM
fifthrapiers
Quick Lin Alg question
If I have the following matrix (3x3):

1 0 1 5
0 1 0 -7/5
0 0 5 7

Is it legal to just take 1/5 of row 3 and say my new matrix is then:

1 0 1 5
0 1 0 -7/5
0 0 1 7/5

? I'm assuming not bc ive checked my work a hundred times and this could be the only possible mistake (Note I am trying to get it in RREF)

EDIT:

Meh, actually I think you CAN... so who wants to help me find out where I went wrong:

ORIG MATRIX:

1 -1 0 5
-1 1 5 2
0 1 1 0

Then switch R_2 and R_3

1 -1 0 5
0 1 1 0
-1 1 5 2

Then R_3 + R_1

1 -1 0 5
0 1 1 0
0 0 5 7

Then R_1 + R_2

1 0 1 5
0 1 1 0
0 0 5 7

Then R_2 - 1/5R_3

1 0 1 5
0 1 0 -7/5
0 0 5 7

Then 1/5R_3

1 0 1 5
0 1 0 -7/5
0 0 1 7/5

Then R_1 - R_3

1 0 0 18/5
0 1 0 -7/5
0 0 1 7/5

No idea
• Sep 4th 2008, 09:37 PM
Chris L T521
Quote:

Originally Posted by fifthrapiers
If I have the following matrix (3x3):

1 0 1 5
0 1 0 -7/5
0 0 5 7

Is it legal to just take 1/5 of row 3 and say my new matrix is then:

1 0 1 5
0 1 0 -7/5
0 0 1 7/5

? I'm assuming not bc ive checked my work a hundred times and this could be the only possible mistake (Note I am trying to get it in RREF)

If you're doing row reduction, that would be legal.

however, you don't have the 3x3 identity matrix appearing in that 3x4 matrix!

to fix that, do this:

$\left[\begin{array}{lllr} 1&0&1&5\\ 0&1&0&-\frac{7}{5}\\ 0&0&1&\frac{7}{5}\end{array}\right]$

$-R_3+R_1\rightarrow R_1$

$\color{red}\boxed{\left[\begin{array}{lllr} 1&0&0&\frac{18}{5}\\ 0&1&0&-\frac{7}{5}\\ 0&0&1&\frac{7}{5}\end{array}\right]}$

I hope this helps! (Sun)

--Chris
• Sep 4th 2008, 09:40 PM
fifthrapiers
Sorry I meant I had a 3x3 matrix with an augmented column. I edited my post. Could you look it over?
• Sep 4th 2008, 09:44 PM
Chris L T521
Quote:

Originally Posted by fifthrapiers
Sorry I meant I had a 3x3 matrix with an augmented column. I edited my post. Could you look it over?

The solution set would be $\left(\tfrac{18}{5},-\tfrac{7}{5},\tfrac{7}{5}\right)$

--Chris
• Sep 4th 2008, 09:46 PM
fifthrapiers
It's not though. The solution is 2, -1, 1.

EDIT: Turns out I copied down the wrong original matrix. No wonder I was spending like an hour trying to figure out what I did wrong. Thanks !
• Sep 4th 2008, 09:49 PM
Chris L T521
When you apply the Gauss-Jordan Method to the augmented matrix, your solution should have the form:

if its a 2x3:

$\left[\begin{array}{llr}1&0&x_0\\ 0&1&y_0\end{array}\right]$

where $\left(x_0,y_0\right)$ is the solution set.

it its a 3x4:

$\left[\begin{array}{lllr}1&0&0&x_0\\ 0&1&0&y_0\\0&0&1&z_0\end{array}\right]$

where $\left(x_0,y_0,z_0\right)$ is the solution set.

I hope this helps you in solving future problems! :D

--Chris