Gauss-Jordan-Elimination-Row

Hi I am learning Gauss Jordan Elimination row but I could not figure out the following question

w + 2x + 2y + 6z = 6

x + y + z = 6

w x + y + 9z = 2

There is an example for me for 2x3 matrix but what is w?

Can someone please illustrate the system in bracket :(?

Your help is very much appriciated

Re: Gauss-Jordan-Elimination-Row

Hello, Newbie999!

Quote:

$\displaystyle \begin{array}{ccc}w + 2x + 2y + 6z &=& 6 \\ \;\;\quad x + y + z &=& 6 \\ w + x + y + 9z &=& 2 \end{array}$

We have: .$\displaystyle \left[\begin{array}{cccc|c}1&2&2&6&6 \\ 0&1&1&1&6 \\ 1&1&1&9&2 \end{array}\right]$

$\displaystyle \begin{array}{c} \\ \\ R_3-R_1\end{array}\left[\begin{array}{cccc|c}1&2&2&6&6 \\ 0&1&1&1&6 \\ 0&\text{-}1&\text{-}1&3&\text{-}4 \end{array}\right]$

$\displaystyle \begin{array}{c}R_1-2R_2 \\ \\ R_3+R_2\end{array}\left[\begin{array}{cccc|c}1&0&0&4&\text{-}6 \\ 0&1&1&1&6 \\ 0&0&0&4&2 \end{array}\right]$

. . . $\displaystyle \begin{array}{c} \\ \\ \frac{1}{4}R_3 \end{array}\left[\begin{array}{cccc|c} 1&0&0&4&\text{-}6 \\ 0&1&1&1&6 \\ 0&0&0&1&\frac{1}{2}\end{array}\right]$

$\displaystyle \begin{array}{c}R_1-4R_3 \\ R_2-R_3 \\ \\ \end{array}\left[\begin{array}{cccc|c}1&0&0&0&\text{-}8 \\ 0&1&1&0&\frac{11}{2} \\ 0&0&0&1 & \frac{1}{2} \end{array}\right]$

We have: .$\displaystyle \begin{Bmatrix}w \;=\;\text{-}8 \\ x+y \;=\;\frac{11}{2} \\ z \;=\;\frac{1}{2}\end{Bmatrix}$

Therefore: .$\displaystyle \begin{Bmatrix} w &=& \text{-}8 \\ x &=& t \\ y &=& \frac{11}{2}-t \\ z &=& \frac{1}{2} \end{Bmatrix}\;\text{ for some value of }t.$

Re: Gauss-Jordan-Elimination-Row

Hi Soroban :) Thank you soo much for giving me the steps.

One more thing, my lecturer actually didn't tell me the reason or the step behind the calculation.

Can you please tell me the theory behind Gauss-Jordan-Elimination and why are the row subtracted or added to one another?

Again, thanks for the help!

Re: Gauss-Jordan-Elimination-Row

You have

w + 2x + 2y + 6z = 6

x + y + z = 6

w x + y + 9z = 2

but Soroban rewrote it as

w + 2x + 2y + 6z = 6

x + y + z = 6

w+ x + y + 9z = 2

is that correct? The "w" is added to x in the last equation?

The row operations are really just a formalized way of reducing the number of unknowns. For example, Soroban subtracted the first row from the third. That is exactly the same as subtracting the two equations, (w+ x+ y+ 9z)- (w + 2x + 2y + 6z)= 2- 6 to eliminate "w":

-x- y+ 3z= -4. Since there is no "w" in the second row, we now have -x- y+ 3z= -4 and x+y+ z= 6. If add those we eliminate **both** x and y leaving only 4z= 2 or z= 1/2. With z= 1/2, x+ y+ z= x+ y+ 1/2= 6 so x+ y= 6- 1/2= 11/2. And we can replace z with 1/2 and y with 11/2- x in the third equation to get w+ x+ 11/2- x+ 9/2= 2. The "x" terms cancel leaving w+ 11/2+ 9/2= w+ 20/2= w+ 10= 2 so w= -8.

That is, x can be any number, y= 11/2- x, w= -8, and z= 1/2.

Re: Gauss-Jordan-Elimination-Row

Thank you guys :)!

I appreciate your time and effort