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Basic Matrix Question

by maherelharake
Tags: basic, matrix
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maherelharake
#1
Jan22-10, 08:27 PM
P: 261
1. The problem statement, all variables and given/known data

Determine, without performing any calculations, whether a linear system with the given augmented matrix has a unique solution, infinitely many solutions, or no solution. Justify your answer.



| 3 -2 0 1 | 1 |
| 1 2 -3 1 | -1|
| 2 4 -6 2 | 0 |


*Not sure how to input matrices into a computer, so sorry for the way I did it.


2. Relevant equations



[b]3. I am not entirely sure how to begin this problem. Any suggestions will be helpful to get me started, and if I still need help I will ask. Please help. Thanks.
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tnutty
#2
Jan22-10, 09:26 PM
P: 328
The key thing is on row3.
maherelharake
#3
Jan22-10, 09:33 PM
P: 261
Thanks for the response! I noticed before I posted my question that the final row had a constant equal to 0. I tried to mentally rearrange the matrix into row echelon form, but I didn't really get much further. I see that it has 4 variables. and 2 nonzero rows. Does that mean that there are 4-2=2 free variables. This is the part that I am getting confused on mostly. Thanks again.

tnutty
#4
Jan22-10, 09:54 PM
P: 328
Basic Matrix Question

Oh wait, there are 3 equation, and 4 variables. Do you think we have enough information
to solve this matrix?
maherelharake
#5
Jan22-10, 09:56 PM
P: 261
I thought about that as well, but I didn't think that choice would be an option according to the directions. :(
Is it possible there is some sort of cancellation after rearranging them mentally? I tried but couldn't get it to work, but it's possible I made a careless error.
tnutty
#6
Jan22-10, 10:15 PM
P: 328
Oh, wait see this :
The original matrix :
[3 -2 0 1 1]
[1 2 -3 1 -1]
[2 4 -6 2 0]

realize that the last row is a multiple of row2. Lets scale the last row3 so we have :

[3 -2 0 1 1]
[1 2 -3 1 -1]
[1 2 -3 1 0]

Notice something there?
maherelharake
#7
Jan22-10, 10:17 PM
P: 261
Does that mean there is no solution since those two equations are the same, and they can't be equal to two different values simultaneously?
maherelharake
#8
Jan22-10, 10:20 PM
P: 261
The left part of the equation is the same, but the right is different. Hence, no solution?
tnutty
#9
Jan22-10, 10:38 PM
P: 328
well try to solve it out and see
maherelharake
#10
Jan22-10, 10:41 PM
P: 261
I will give it a shot, but the problem doesn't want me to solve it out. I will try it anyways and repost when I give it my best attempt.
tnutty
#11
Jan22-10, 10:44 PM
P: 328
Its ok, the problem is trying to make you think hard, but I think once you thought about
the problem a little, you should then confirm it by solving the matrix.
maherelharake
#12
Jan22-10, 10:56 PM
P: 261
When I try to solve it I come up with
[1 -2/3 0 1/3 1/3]
[0 0 0 0 -1]
[0 0 0 0 1]

that means that x1=x2=x3=x4=0, however using these values, it is not possible to attain both 1 and -1 simultaneously. Thus the answer is no solution.

Did I make a mistake along the way?
Dick
#13
Jan22-10, 11:01 PM
Sci Advisor
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Thanks
P: 25,247
Quote Quote by maherelharake View Post
Does that mean there is no solution since those two equations are the same, and they can't be equal to two different values simultaneously?
Yes. [1 2 -3 1 -1] and [1 2 -3 1 0] and translating them into equations, they can't be true simultaneously. Why are we dragging this out?
tnutty
#14
Jan22-10, 11:03 PM
P: 328
In the last row you have this : [0 0 0 0 1] which means this in equation form,
0X1 + 0X2 + 0X3 + 0X4 = 1, which means 0 = 1. This is impossible mathematically, right?
Because a 0 cannot never equal 1, which means that the system is inconsistent, or no solution. So Yes you are correct as to your final conclusion.
maherelharake
#15
Jan22-10, 11:18 PM
P: 261
Ok guys thanks a lot. I really appreciate it. I'm new to all of this stuff and I am still getting my feet under me. Thanks again.
HallsofIvy
#16
Jan23-10, 08:51 AM
Math
Emeritus
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Thanks
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P: 39,345
Quote Quote by tnutty View Post
Oh, wait see this :
The original matrix :
[3 -2 0 1 1]
[1 2 -3 1 -1]
[2 4 -6 2 0]
This is the crucial point, and why you don't need to do any calculation. All numbers in the third row are two times the numbers in the second row except the last. That tells us there is NO solution. If all numbers in the third row had been a multiple of the second row (and first and second rows were independent) there would have been an infinite number of solutions.

realize that the last row is a multiple of row2. Lets scale the last row3 so we have :

[3 -2 0 1 1]
[1 2 -3 1 -1]
[1 2 -3 1 0]

Notice something there?
maherelharake
#17
Jan23-10, 11:16 AM
P: 261
Thanks a lot. That clear explanation really helped out even more.


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