# Gauss-Jordan

• Jan 23rd 2009, 09:46 AM
luxdelux
Gauss-Jordan
Hi again, I'm not sure if it's okay for me to post in here again, as I posted last night--but this is truly urgent. I'll be taking a midterm in a short while and there are several questions on my study guide that have me completely stumped:

1. y + z = 2
x + y + z = -1
-x + z = -1

I have done this problem again and again, only to end up with different/wrong answers for x, y, and z each time. I know that I have to created an augmented matrix, so I did that, and then I continued to row reduce until I got the identity matrix on my left, and the the supposed (yet incorrect) solutions on the right. I don't know what I'm doing wrong here, and I'm beyond frustrated.

2. 2y + z = 0
y + z = 1
x - y = -1

Same situation here. I created the augmented matrix and then, when I saw that column one is [0 0 1], I don't know what to do.

3. -x + y +2z = 1
2x -2z = 0
2x + y + 2z = 0

I switched the first and second row here, then divided the first row by 2, and proceeded to use G-J, but messed up my calculations because I'm not getting the correct solutions for x, y, and z...I've re-done over and over and I'm not sure what I'm doing incorrectly.

4. 2x + y - z = 0
-x + y = 1
2x + y + 2z = 0

Same situation here...I've done the problem repeatedly only to end up with different x, y and z each time.

5. Suppose a two by two matrix A is nonsingular and is defined as follows:

A = ( a b
0 d ) < The whole thing is supposed to be in the parenthesis

where a, b, d are real numbers. Use Gauss-Jordan to find the inverse of A.

Again...I am at a total loss here. I think that I have to put an identity matrix in there somewhere?

I'd really appreciate the help on this...I have no idea what I'm doing, and I'm definitely panicking. THANK YOU!
• Jan 23rd 2009, 10:22 AM
Soroban
Hello, luxdelux!

Quote:

$\displaystyle (1)\;\;\begin{array}{ccc}\quad\;\; y + z &=& 2 \\ x + y + z &=& \text{-}1 \\ \text{-}x \quad\;\; + z &=& \text{-}1 \end{array}$

Here's something that obviously didn't occur to you.

I'm not laughing at you ... It took me a long time to realize this.

We can re-order the equations: .$\displaystyle \begin{array}{ccc}x + y + z &=& \text{-}1 \\ \qquad y + z &=& 2 \\ \text{-}x \quad\;\;+z &=& \text{-}1 \end{array}$

We have: .$\displaystyle \left[\begin{array}{ccc|c}1&1&1 & \text{-}1 \\ 0 &1&1&2 \\ \text{-}1 &0&1&\text{-}1 \end{array}\right]$

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

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

. . $\displaystyle \begin{array}{c}\\ R_2-R_3\\ \end{array}\left[\begin{array}{ccc|c}1&0&0&\text{-}3 \\ 0&1&0&6 \\ 0&0&1&\text{-}4 \end{array}\right]$

Therefore: .$\displaystyle \begin{Bmatrix}x &=& \text{-}3 \\ y &=& 6 \\ z &=& \text{-}4 \end{Bmatrix}$