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Basic Matrix Question 
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#1
Jan2210, 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. 


#2
Jan2210, 09:26 PM

P: 328

The key thing is on row3.



#3
Jan2210, 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 42=2 free variables. This is the part that I am getting confused on mostly. Thanks again.



#4
Jan2210, 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? 


#5
Jan2210, 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. 


#6
Jan2210, 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? 


#7
Jan2210, 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?



#8
Jan2210, 10:20 PM

P: 261

The left part of the equation is the same, but the right is different. Hence, no solution?



#9
Jan2210, 10:38 PM

P: 328

well try to solve it out and see



#10
Jan2210, 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.



#11
Jan2210, 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. 


#12
Jan2210, 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? 


#13
Jan2210, 11:01 PM

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#14
Jan2210, 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. 


#15
Jan2210, 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.



#16
Jan2310, 08:51 AM

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#17
Jan2310, 11:16 AM

P: 261

Thanks a lot. That clear explanation really helped out even more.



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