# Thread: Solve using Gauss–Jordan method...?

1. ## Solve using Gauss–Jordan method...?

Use the Gauss–Jordan method to solve the following system of linear equations:

(A system with more variable than equations is called
underdetermined.)

x
+ y z + 2w = 20
2
x y + z + w = 11

3
x 2y + z 2w = 27

So putting this into matrices, I get
[1 1 -1 2 | -20]
[2 -1 1 1 | 11 ]
[3 -2 1 -2 | 27]
^ note, that's all supposed to be in one matrix

I've tried over and over again, but I just can't get this system into row-reduced form...can someone please help?

2. Originally Posted by tuheetuhee
Use the Gauss–Jordan method to solve the following system of linear equations:

(A system with more variable than equations is called
underdetermined.)
x
+ y z + 2w = 20
2
x y + z + w = 11

3
x 2y + z 2w = 27

So putting this into matrices, I get
[1 1 -1 2 | -20]
[2 -1 1 1 | 11 ]
[3 -2 1 -2 | 27]
^ note, that's all supposed to be in one matrix

I've tried over and over again, but I just can't get this system into row-reduced form...can someone please help?
R1, R2, and R3 stands for row 1, row 2, and row 3, respectively, in the previous matrix

$\begin{array}{cccc|cl} 1 & 1 & -1 & 2 & -20 & \\
2 & -1 & 1 & 1 & 11 & \\
3 & -2 & 1 & -2 & 27 & \\
\hline 1 & 1 & -1 & 2 & -20 & \text{.......... R1} \\
0 & 3 & -3 & 3 & -51 & \text{.......... 2R1 - R2} \\
0 & 5 & -4 & 8 & -87 & \text{.......... 3R1 - R3} \end{array}$

$\begin{array}{cccc|cl} \hline 1 & ~{\color{white}..}0 & 0 & 3 & -9 & \text{.......... (3R1 - R2) / 3} \\
0 & 1 & -1 & 1 & -17 & \text{.......... R2 / 3} \\
0 & 0 & -1 & -3 & 2 & \text{.......... (5/3)R2 - R3} \\
\hline 1 & 0 & 0 & 3 & -9 & \text{.......... R1} \\
0 & 1 & 0 & 4 & -19 & \text{.......... R2 - R3} \\
0 & 0 & 1 & 3 & -2 & \text{.......... -R3} \\
\hline \end{array}$

3. Hello, tuheetuhee!

Use the Gauss–Jordan method to solve the following system of linear equations:

. . $\begin{array}{ccc}x + y - z + 2w &=& \text{-}20 \\ 2x - y + z + w &= &11 \\ 3x - 2y + z - 2w &=& 27 \end{array}$

We have: . $\left[\begin{array}{cccc|c}1 & 1 & \text{-}1 & 2 & \text{-}20 \\ 2 & \text{-}1 & 1 & 1 & 11 \\ 3 & \text{-}2 & 1 & \text{-}2 & 27 \end{array}\right]$

$\begin{array}{c} \\ R_2-2R_2 \\ R_3 - 3R_1 \end{array}\left[\begin{array}{cccc|c}1 & 1 & \text{-}1 & 2 & \text{-}20 \\ 0 & \text{-}3 & 3 & \text{-}3 & 51 \\ 0 & \text{-}5 & 4 & \text{-}8 & 87 \end{array}\right]$

. . $\begin{array}{c} \\ -\frac{1}{3}R_2 \\ \\ \end{array}\left[\begin{array}{cccc|c} 1 & 1 & \text{-}1 & 2 & \text{-}20 \\ 0 & 1 & \text{-}1 & 1 & \text{-}17 \\ 0 & \text{-}5 & 4 & \text{-}8 & 87 \end{array}\right]$

$\begin{array}{c}R_1-R_2 \\ \\ R_3 + 5R_2\end{array} \left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & \text{-}3 \\ 0 & 1 & \text{-}1 & 1 & \text{-}17 \\ 0 & 0 & \text{-}1 & \text{-}3 & 2 \end{array}\right]$

$\begin{array}{c} \\ R_2-R_3 \\ \text{-}1\!\cdot\!R_3\end{array}\left[\begin{array}{cccc|c}1 & 0 & 0 & 1 & \text{-}3 \\ 0 & 1 & 0 & 4 & \text{-}19 \\ 0 & 0 & 1 & 3 & \text{-}2\end{array}\right]$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: . $\begin{array}{ccc}x + w &=& \text{-}3 \\ y + 4w &=&\text{-}19 \\ z + 3w &=& \text{-}2 \end{array}\quad\Rightarrow\quad \begin{array}{ccc}x &=& \text{-} 3 - w \\ y &=& \text{-}19 - 4w \\ z &=& \text{-}2 - 3w \\ w &=& w \end{array}$

On the right side, replace $w$ with the parameter $t$.

Therefore: . $\boxed{\begin{array}{ccc}x &=& \text{-}3 - t \\ y &=& \text{-}19 - 4t \\ z &=& \text{-}2 - 3t \\ w &=& t \end{array} \quad\text{ for any real number }t\;\; }$

This represents all the possible solutions of the system.

4. Thank you Jhevon and Soroban!
Your solutions really make sense.Thank you so much for your help! Really appreciate it.

By the way, so for underdetermined systems of equations, we don't have to bring the matric to row-reduced form then?

5. Originally Posted by tuheetuhee
Thank you Jhevon and Soroban!
Your solutions really make sense.Thank you so much for your help! Really appreciate it.

By the way, so for underdetermined systems of equations, we don't have to bring the matric to row-reduced form then?
we put the matrix into reduced row-echelon form here, but no, you do not have to do that when solving any system of equations. it is enough to bring it to row echelon form (the form where all non-zero rows have leading 1's). the difference is, the algebra at the end that you need to do to find the solutions might be a bit more tedious. unless you are told specifically what to do, it is up to you whether you want to work out the algebra or just bring it to reduced row echelon form and do easier algebra.

6. Oh okay! That makes sense...you put teh matrix into row-reduced form when you want to make the algebra easier. Thanks a lot for your help Jhevon!

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