gaussian elimination

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July 9th 2006, 02:07 AM
cheesepie

need help urgent !!

Gaussin Elimation

1)x+2y+6z=12
2z=2
-3x-6y-9z=-27

2)x-y+z=22
2y-z=-23
2x+3y=-11.2

3)x+2y+4z=7
3x+y+4z=-2
2x+9y-2x=10

4)x+2y+4z=7
3x+y+4z=-2
2x+9y-2z=10
July 9th 2006, 06:37 AM
Soroban

Hello, cheesepie!

#1 is a set of dependent equations.

Quote:

$1)\;\begin{array}{ccc}x+2y+6z\:=\:12\\2z\:=\:2 \\ -3x-6y-9z\:=\:-27\end{array}$

We have: . $\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 2 & | & 2 \\ \text{-}3 & \text{-}6 & \text{-}9 & | & \text{-}27\end{vmatrix}$

$\begin{array}{ccc} \\ R_2\div2\\ R_3\div(\text{-}3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 1 & 2 & 3 & | & 9\end{vmatrix}$

$\begin{array}{ccc}\\ \\ R_3-R_1\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & \text{-}3 & | & \text{-}3\end{vmatrix}$

$\begin{array}{ccc} \\ \\ R_3\div(-3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 1 & | & 1\end{vmatrix}$

$\begin{array}{ccc} \\ \\ R_3-R_2\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0\end{vmatrix}$

Since we have a rows of 0's (zeros all the way across), the system is dependent.

There is an infinite number of solutions to the system.
July 10th 2006, 05:34 AM
cheesepie

tks...how about other question ? :D
July 10th 2006, 07:06 AM
Soroban

Hello, cheesepie1

I need to know exactly where your difficulty is . . .
. . You don't understand Gaussian Elimination at all?
. . You don't know how to get the 1's and 0's?
. . You don't know what order to work at it?

I can spend hours, doing ALL your homework for you,
. . but there is no learning taking place, it's wasted effort.

Having said that, here's the second one.
And I would really like to know exactly why you can't do the third one.

Quote:

$2)\;\begin{array}{ccc}x-y+z \:=\:22 \\ 2y - z\:=\:-23 \\ 2x + 3y\:=\:-11.2\end{array}$

We have: . $\begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 2 & \text{-}1 & | & \text{-}23 \\ 2 & 3 & 0 & | & \text{-}11.2\end{vmatrix}$

$\begin{array}{ccc} \\R_2\div2\\R_3-2\cdot R_1\end{array}\;\begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 5 & \text{-}2 & | & \text{-}55.2\end{vmatrix}$

$\begin{array}{ccc}R_1+R_2\\ \\ R_3-5\cdotR_2\end{array}\;\begin{vmatrix}1 & 0 & 0.5 & | & 10.5 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 0 & 0.5 & | & 2.3\end{vmatrix}$

$\begin{array}{cccc} R_1-0.5\cdot R_3 \\ R_2+0.5\cdot R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & 8.2 \\ 0 & 1 & 0 & | & \text{-}9.2 \\ 0 & 0 & 1 & | & 4.6\end{vmatrix}$

Answers: . $x = 8.2,\;y = -9.2,\;z = 4.6$

And remember, you can check your answers yourself.
July 10th 2006, 07:15 AM
cheesepie

i dont understand the inverse matrices

example how u change inverse of b =
[ 8 4 ]
[ 3 2 ]

into

[1 0 ]
[0 1 ]
July 10th 2006, 07:56 AM
topsquark

You aren't inverting matrices using this method. You are performing "row operations." For example, you have the system of equations:
(1) x - y + z = 22
(2) 2y - z = -23
(3) 2x + 3y = -11.2

I will solve this in an identical manner as Soroban, except that he is using Gaussian elimination and I am simply manipulating equations. What I want you to note is that they are the same thing, just that one is in matrix notation and one is not.

So. The first operation Soroban did was R2 $\div$ 2. This is the same as dividing both sides of equation 2 by 2:

(1) x - y + z = 22
(2)' y - 0.5z = -11.5
(3) 2x + 3y = -11.2

The next thing he did was R3 - 2R1. This is subtracting twice equation 1 by R3.
2x + 3y = -11.2 (Equation 3)
-2x + 2y - 2z = -44 (Equation 1 with both sides multiplied by -2 )
---------------------------
(3)' 5y - 2z = -55.2

So we have:
(1) x - y + z = 22
(2)' y - 0.5z = -11.5
(3)' 5y - 2z = -55.2

Now R1 + R2. This is equation 1 + equation 2'

x - y + z = 22
y -0.5z = -11.5
-------------------
(1)' x + 0.5z = 10.5

So
(1)' x + 0.5z = 10.5
(2)' y - 0.5z = -11.5
(3)' 5y - 2z = -55.2

Now R3 - 5R2 means equation 3' - 5 times equation 2"
5y - 2z = -55.2
-5y +2.5z = 57.5
----------------------
(3)" 0.5z = 2.3

So we have

(1)' x + 0.5z = 10.5
(2)' y - 0.5z = -11.5
(3)" 0.5z = 2.3

R1 - R3 means equation 1' minus 0.5 times equation 3" (Soroban's statement here was incorrect.)
x + 0.5z = 10.5
- 0.5z = -2.3
----------------
(1)" x = 8.2

and R2 + R3 means equation 2' plus equation 3" (Soroban's statement here was incorrect.)
y - 0.5z = -11.5
0.5z = 2.3
----------------
(2)"' y = -9.2

So
(1)" x = 8.2
(2)"' y = -9.2
(3)" 0.5z = 2.3 (Then, of course, solve this for z!)

You can compare each step in the above with Soroban's matrix form.

-Dan
July 10th 2006, 11:02 AM
Soroban

Hello, cheesepie!

Now you've changed the subject . . .

Quote:

i dont understand the inverse matrices

Example: how u change inverse of b =
[ 8 4 ]
[ 3 2 ]

into

[1 0 ]
[0 1 ]

You haven't been taught this procedure either??

The set up is: / $\begin{bmatrix}8 & 4 & | & 1 & 0 \\ 3 & 2 & | & 0 & 1\end{bmatrix}$

Now, use Gaussian Elimination to change the left matrix into the Identity: . $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$