1. ## 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!

2. Hello, luxdelux!

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

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: . $\begin{array}{ccc}x + y + z &=& \text{-}1 \\ \qquad y + z &=& 2 \\ \text{-}x \quad\;\;+z &=& \text{-}1 \end{array}$

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

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

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

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

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