Finite Math

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Tiger-T

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40.) 4x – 3y + z + w =21, -2x – y + 2z + 7w = 2, and 10x – 5z – 20w =15


4 3 1 1 ? 21
-2 -1 2 7 ? 2
10 0 5 20 ? 15

this is to be solved the Gauss-Jordan method. The book and the online turtorials do not explain how to do decimals. The decimals appear when I divide the first row by the first number in the first column 4. Or would I beging by making the -2 in the second row first column a 0? The book states that I should start with a 1 in the first row and column, but as the process continues the 1 changes to other numbers. I am confused on how to get started. Please help!!!
 
Tiger-T said:
40.) 4x – 3y + z + w =21, -2x – y + 2z + 7w = 2, and 10x – 5z – 20w =15


4 3 1 1 ? 21
-2 -1 2 7 ? 2
10 0 5 20 ? 15

this is to be solved the Gauss-Jordan method. The book and the online turtorials do not explain how to do decimals. The decimals appear when I divide the first row by the first number in the first column 4. Or would I beging by making the -2 in the second row first column a 0? The book states that I should start with a 1 in the first row and column, but as the process continues the 1 changes to other numbers. I am confused on how to get started. Please help!!!
Click to expand...
You've got 3 equations only, but 4 variables :shock:

Your top line should be: 4 -3 1 1 | 21
and third line should be: 10 0 -5 -20 | 15

I can't "follow" what you're saying...

Please post the initial exercise IN FULL.
 
Tiger-T said:
The book and the online turtorials do not explain how to do decimals.

You could use Rational numbers, instead.
Click to expand...

Code: 
The result of "multiply Row 1 by 1/4" is:

                              |
1    -3/4     1/4     1/4     |     21/4
                              |



The result of "add two times Row 1 to Row 2" is:

                              |
0    -5/2     5/2     15/2    |     25/2
                              |

Et cetera

 
I am afraid that this is the entire problem. Not much to go on huh? I am just supposes to solve it. I am stuck on where to even begin.
 
Hello, Tiger-T!

\(\displaystyle 40)\;\;\begin{array}{cccccc}4x - 3y + z + w &=& 21 & [1] \\ -2x - y + 2z + 7w &=& 2 & [2] \\ 10x\quad\; - 5z - 20w &=& 15 & [3] \end{array}\)
Click to expand...

There are many things your book did not tell you.
. . Do NOT introduce fraction or decimals (unless you're forced to).
. . We can change the order of the equations if we want to.

Note that equation [3] can be divided through by 5: .\(\displaystyle 2x - z - 4w \:=\:3\)
. . I'll write this equation on top.
. . and write equation [1] on the bottom.


\(\displaystyle \text{We have: }\;\left|\begin{array}{cccc|c}2 & 0 & \text{-}1 & \text{-}4 & 3 \\ \text{-}2 & \text{-}1 & 2 & 7 & 2 \\ 4 & \text{-}3 & 1 & 1 & 21 \end{array}\right|\)


\(\displaystyle \begin{array}{c} \\ R_2 + R_1 \\ R_3 - 2R_1 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & \text{-}4 & 3 \\ 0 & \text{-}1 & 1 & 3 & 5 \\ 0 & \text{-}3 & 3 & 9 & 15 \end{array}\right|\)


. . . \(\displaystyle \begin{array}{c} \\ \text{-}1\!\cdot\!R_2 \\ \text{-}\frac{1}{3}\!\cdot\!R_3 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & 4 & 3 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \end{array}\right|\)


. \(\displaystyle \begin{array}{c} \\ \\ R_3-R_2 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & 4 & 3 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|\)


. . . \(\displaystyle \begin{array}{c}\frac{1}{2}\!\cdot\!R_1 \\ \\ \\ \end{array} \left|\begin{array}{cccc|c} 1 & 0 & \text{-}\frac{1}{2} & 2 & \frac{3}{2} \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|\)


We have an infinite number of possible solutions.
(But we knew that at the very beginning, didn't we?)


\(\displaystyle \text{We have: }\;\begin{Bmatrix} x - \frac{1}{2}z + 2w &=& \frac{3}{2} \\ y - z - 3w &=& \text{-}5 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix} x &=& \frac{3}{2} + \frac{1}{2}z - 2w \\ y &=& \text{-}5 + z - 3w \\ z &=& z \\ w &=& w \end{Bmatrix}\)


\(\displaystyle \text{On the right, replace }z\text{ with }t,\,\text{ and }w\text{ with }u.\)


\(\displaystyle \text{We have a set of parametric equations for all solutions of the system of equations.}\)

. . . . . \(\displaystyle \begin{Bmatrix}x &=& \frac{3}{2} + \frac{1}{2}t - 2u \\ y &=& \text{-}5 + t + 3u \\ z &=& t \\ w &=& u \end{Bmatrix}\)

 
Tiger-T said:
4x – 3y + z + w =21, -2x – y + 2z + 7w = 2, and 10x – 5z – 20w =15
Click to expand...
4x – 3y + z + w =21 [1]
-2x – y + 2z + 7w = 2 [2]
10x – 5z – 20w =15 [3]

4x – 3y + z + w =21 [1]
6x + 3y -6z - 21w = -6 [2] times -3
Add these to get:
10x -5z - 20w = 15 : SAME as equation [3] !

Do you see now why the problem makes no sense? :shock:
 
Soroban,
I cannot thank you enough!!!!



soroban said:
Hello, Tiger-T!

\(\displaystyle 40)\;\;\begin{array}{cccccc}4x - 3y + z + w &=& 21 & [1] \\ -2x - y + 2z + 7w &=& 2 & [2] \\ 10x\quad\; - 5z - 20w &=& 15 & [3] \end{array}\)
Click to expand...

There are many things your book did not tell you.
. . Do NOT introduce fraction or decimals (unless you're forced to).
. . We can change the order of the equations if we want to.

Note that equation [3] can be divided through by 5: .\(\displaystyle 2x - z - 4w \:=\:3\)
. . I'll write this equation on top.
. . and write equation [1] on the bottom.


\(\displaystyle \text{We have: }\;\left|\begin{array}{cccc|c}2 & 0 & \text{-}1 & \text{-}4 & 3 \\ \text{-}2 & \text{-}1 & 2 & 7 & 2 \\ 4 & \text{-}3 & 1 & 1 & 21 \end{array}\right|\)


\(\displaystyle \begin{array}{c} \\ R_2 + R_1 \\ R_3 - 2R_1 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & \text{-}4 & 3 \\ 0 & \text{-}1 & 1 & 3 & 5 \\ 0 & \text{-}3 & 3 & 9 & 15 \end{array}\right|\)


. . . \(\displaystyle \begin{array}{c} \\ \text{-}1\!\cdot\!R_2 \\ \text{-}\frac{1}{3}\!\cdot\!R_3 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & 4 & 3 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \end{array}\right|\)


. \(\displaystyle \begin{array}{c} \\ \\ R_3-R_2 \end{array} \left|\begin{array}{cccc|c} 2 & 0 & \text{-}1 & 4 & 3 \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|\)


. . . \(\displaystyle \begin{array}{c}\frac{1}{2}\!\cdot\!R_1 \\ \\ \\ \end{array} \left|\begin{array}{cccc|c} 1 & 0 & \text{-}\frac{1}{2} & 2 & \frac{3}{2} \\ 0 & 1 & \text{-}1 & \text{-}3 & \text{-}5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|\)


We have an infinite number of possible solutions.
(But we knew that at the very beginning, didn't we?)


\(\displaystyle \text{We have: }\;\begin{Bmatrix} x - \frac{1}{2}z + 2w &=& \frac{3}{2} \\ y - z - 3w &=& \text{-}5 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix} x &=& \frac{3}{2} + \frac{1}{2}z - 2w \\ y &=& \text{-}5 + z - 3w \\ z &=& z \\ w &=& w \end{Bmatrix}\)


\(\displaystyle \text{On the right, replace }z\text{ with }t,\,\text{ and }w\text{ with }u.\)


\(\displaystyle \text{We have a set of parametric equations for all solutions of the system of equations.}\)

. . . . . \(\displaystyle \begin{Bmatrix}x &=& \frac{3}{2} + \frac{1}{2}t - 2u \\ y &=& \text{-}5 + t + 3u \\ z &=& t \\ w &=& u \end{Bmatrix}\)
Click to expand...
 
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