# Basic Matrix Question

by maherelharake
Tags: basic, matrix
 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.
 P: 328 The key thing is on row3.
 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.
 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?
 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.
 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?
 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?
 P: 261 The left part of the equation is the same, but the right is different. Hence, no solution?
 P: 328 well try to solve it out and see
 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.
 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.
 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?
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P: 25,228
 Quote by maherelharake 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?
 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.
 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.
Math
Emeritus