gaussian elimiation with out pivoting - help!!!

Printable View

November 23rd 2012, 04:17 AM
kaya2345

1 Attachment(s)

gaussian elimiation with out pivoting - help!!!

I am stuck on a question that i'm doing on gaussian elimination.

Any help is appreciated!
November 23rd 2012, 06:29 AM
Soroban

Re: gaussian elimiation with out pivoting - help!!!

Hello, kaya2345!

Quote:

Solve the following system of equations using Gaussian Elimination without pivots.
Not sure what that means . . .

. . $\begin{array}{ccc}6x_1+7x_2+8x_3 \:=\:21 \\ 7x_1+8x_2+9x_3 \:=\:24 \\ 8x_1+9x_2+9x_3 \:=\:26 \end{array}$

We have: . $\left[\begin{array}{ccc|c} 6&7&8&21 \\ 7&8&9&24 \\ 8&9&9&26 \end{array}\right]$

$\begin{array}{c}\\ R_2-R_1 \\ R_3-R_2 \end{array}\left[\begin{array}{ccc|c} 6&7&8&21 \\ 1&1&1&3 \\ 1&1&0&2 \end{array}\right]$

$\begin{array}{c}R_1-6R_2 \\ R_2-R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}0&1&2&3 \\ 0&0&1&1 \\ 1&1&0&2 \end{array}\right]$

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

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

Therefore: . $\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} \;=\;\begin{bmatrix}1\\1\\1 \end{bmatrix}$

What's that nonsense about four decimal places?!
And since the roots are equal,
. . who cares where the pivots are?
November 23rd 2012, 11:30 AM
BobP

Re: gaussian elimiation with out pivoting - help!!!

This is an examination or coursework question so you have to be aware that you are being tested on the use of a particular method, not on whether you can simply solve the system. You might use some method that gets you the exact solution, but if it isn't the method being asked for you will earn no marks for your solution.
Gaussian elimination without any pivoting requires you to take multiples of the first equation from the second and third equations to so as to eliminate the first of the unknowns from those equations. Those multiples will be 7/6 and 8/6, and you work to the stated degree of accuracy. You then take a multiple of the new second equation from the third equation so as to remove the second of the unknowns. The third of the unknowns can then be calculated and back substitution gets you the other two. The results will naturally be approximations to the exact values.
November 24th 2012, 12:15 PM
kaya2345

Re: gaussian elimiation with out pivoting - help!!!

Quote:

Originally Posted by BobP

This is an examination or coursework question so you have to be aware that you are being tested on the use of a particular method, not on whether you can simply solve the system. You might use some method that gets you the exact solution, but if it isn't the method being asked for you will earn no marks for your solution.
Gaussian elimination without any pivoting requires you to take multiples of the first equation from the second and third equations to so as to eliminate the first of the unknowns from those equations. Those multiples will be 7/6 and 8/6, and you work to the stated degree of accuracy. You then take a multiple of the new second equation from the third equation so as to remove the second of the unknowns. The third of the unknowns can then be calculated and back substitution gets you the other two. The results will naturally be approximations to the exact values.

Thank you for that! my values for x1, 2 and 3 were very close to 1 so I assume that is right?

Also is there any chance you could shed some light on the vector segment of the question?
November 25th 2012, 01:50 AM
BobP

Re: gaussian elimiation with out pivoting - help!!!

Just substitute the values you've calculated into each equation and find the difference with the rhs in each case. Those are the residuals.
For the last past read your notes on LU factorisation.