Solved

# gauss elimination with parameters

Posted on 2008-11-19
Medium Priority
449 Views
Last Modified: 2016-02-10
I'm breaking my head trying to figure out what i'm doing wrong,
i'm support to eliminate the following equations using Gauss elimination system

ax+y+z=4
x+by+z=3
x+2by+z=4
and i'm suppose to figure out for which a,b (which are Parameters) the system has One solution/Infinit Solutions/No solutions - if there's infint solutions i'm suppose to write the general solution for the system.
now it's all pretty basic in the begining

1    b  1  3
1  2b  1  4 R2->R2-R1
a    1  1  4 R3->R3-aR1

which leads me to

1   b           1  3
0   b           0  1       * i can reach the first value using /:b , i then get y=1/b
0  1-ab   1-a  4-3a

from here i'm puzzled as there's no way to carry on the elimination (if i'm doing it right)
0
Question by:d3bugg3r
• 3
• 2
6 Comments

LVL 27

Expert Comment

ID: 22998098
You are making things too complicated.

What happens if   a = b = 1 ??

What values will make the 1st and 3rd equations wierd??
0

LVL 3

Author Comment

ID: 22998220
I dont see how it's possible, it will give the following result for the 3rs equation:
0 0 0 0 1 which is not possible, not as a solution anyway.
0

LVL 27

Expert Comment

ID: 22998258
What happens to the 1st and 2nd equations if   a = b = 1 ??
You don't have to solve them, just look at them.

0

LVL 3

Author Comment

ID: 22998415
yes, it works out... but it doesnt for the 3rd so i cannot use those values as answer.
and i have to define if for those two parameters there's uniqe solution, unlimited solutions or no solutions (like if b=0 then 1/0 on the second equation isnt possible)
0

LVL 27

Accepted Solution

d-glitch earned 500 total points
ID: 22998554
There are three cases to consider:

1st and 2nd    ==>   a = b = 1    ==>  No solutions

1st and 3rd    ==>   a=1  b=0.5  ==>  Infinite solutions

2nd and 3rd   ==>  b=0 (which you found)  ==>  No solutions

I'm pretty sure that all other cases have single solutions.
0

LVL 22

Assisted Solution

NovaDenizen earned 500 total points
ID: 23005425
Whenever you have an equation involving a product like
r * s = t

When t is zero, at least one of r and s must be zero
When t is nonzero, both r and s must be nonzero

Also, whenever you want to eliminate a factor like in
a*b = 1
to get
a = 1/b
You simply cannot do that unless you have proven that b is nonzero.

0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Complex Numbers are funny things.  Many people have a basic understanding of them, some a more advanced.  The confusion usually arises when that pesky i (or j for Electrical Engineers) appears and understanding the meaning of a square root of a nega…
Lithium-ion batteries area cornerstone of today's portable electronic devices, and even though they are relied upon heavily, their chemistry and origin are not of common knowledge. This article is about a device on which every smartphone, laptop, an…
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac…
I've attached the XLSM Excel spreadsheet I used in the video and also text files containing the macros used below. https://filedb.experts-exchange.com/incoming/2017/03_w12/1151775/Permutations.txt https://filedb.experts-exchange.com/incoming/201…
###### Suggested Courses
Course of the Month16 days, 5 hours left to enroll

#### 850 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.