HELP WITH LINEAR SYSTEM QUESTION
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D
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You do not have an equation with "x2". Is that intentional or over site?
That is giving me the question, of which some parts I have answered and other parts I am unsure on. If you know how to help, then just help, if you cannot help or do not want to then please refrain from replying.
D
Deleted member 4993
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You have NOT shared your work or thoughts regarding your assignment.
Please show your work - telling us how you arrived at the answers that you provided. That will tell us where to begin to help you.
In particular, please tell us the mathematical definition of "Consistent System".
Steven G
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I think that you and the helpers on this forum have a different understanding of what help means. It seems that your idea of help is for us to give you the answer. We do not do that here. If that is unsatisfactory that is fine but then you need to go elsewhere for help. The helpers here want you to do the problem with helpful hints from us. Before you can do this problem you need to know what information you are given and what it means.
I prefer that you stay around here but if you do then you need to answer my question. Even if it is not a perfect answer we can help you understand what information you are given.
Steven G
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Thank you!
A consistent system is a system that has at least one solution, either a unique solution or infinite number of solutions. 8 equations for n unknowns implies a matrix with 8 rows and n columns. The rank of the matrix is 5 which means there are 5 leading ones so 5 restrictions on n variables
bi) The minimum number of n unknowns is 5 as there are 5 leading ones in the solution, so there is at least 5 leading variables.
bii) If b = 0, it will have the trivial solution of x=0 and therefore be consistent? This is the one I am most unsure of
biii) If b =/=0 it may not be consistent as the matrix does not have full row rank, which guarantees consistency, so there could possibly be 3 inconsistent rows of the matrix if b is not zero. looking of the form for example 0x + 0y + 0z = 13. So the system is not consistent for all values of b, the only value to guarantee consistency is b=0 as we know there will be no inconsistent rows. The system would only be consistent regardless of the value of b if the system had full row rank, which is not the case here as the rank is 5 and there are 8 rows in total.
biv)If the linear system is consistent, the minimum number of variables to guarantee an infinite number of solutions is 6. this is because there are 5 leading ones which implies there is 5 leading variables,if just 5 variables we would have a single point in R^5 but having an extra variable that is a free parameter means there will be a complete line of solutions and hence infinite number of solutions
bv)A four dimensional flat in R^n implies there are 4 free parameters. As we already know there are 5 leading variables that are not free, the value of n must be 9 (5+4)
Hope this helps
Is my reasoning sound ? Especially for bii)
Thanks in advance
bi) The minimum number of n unknowns is 5 as there are 5 leading ones in the solution, so there is at least 5 leading variables.
bii) If b = 0, it will have the trivial solution of x=0 and therefore be consistent? This is the one I am most unsure of
biii) If b =/=0 it may not be consistent as the matrix does not have full row rank, which guarantees consistency, so there could possibly be 3 inconsistent rows of the matrix if b is not zero. looking of the form for example 0x + 0y + 0z = 13. So the system is not consistent for all values of b, the only value to guarantee consistency is b=0 as we know there will be no inconsistent rows. The system would only be consistent regardless of the value of b if the system had full row rank, which is not the case here as the rank is 5 and there are 8 rows in total.
biv)If the linear system is consistent, the minimum number of variables to guarantee an infinite number of solutions is 6. this is because there are 5 leading ones which implies there is 5 leading variables,if just 5 variables we would have a single point in R^5 but having an extra variable that is a free parameter means there will be a complete line of solutions and hence infinite number of solutions
bv)A four dimensional flat in R^n implies there are 4 free parameters. As we already know there are 5 leading variables that are not free, the value of n must be 9 (5+4)
Hope this helps
Is my reasoning sound ? Especially for bii)
Thanks in advance
Hmm ... a consistent system. If you are solving an equation Ax = b, and we're staying in one space (what does that mean?) then we want a square matrix A. In this case, we're talking about 8 rows and 8 columns. To solve the equation Ax = b, we use (at least I use) an augmented matrix with 8 rows but 9 columns. The b (target vector) will be the 9th column. Then we row reduce to RREF. Only in the RREF (row-reduced echelon form, I assume you know what that means?) will you see those leading 1's. If the rank of A is 5, there will be 5 leading 1's, in the RREF that is. When you have the RREF, how do you know whether the system is consistent or not?
If b = the 0 vector, then yes, the system is consistent, but the only solution is the trivial one. If the rank is 8 (in this case, or n in the general case) that will be the only solution. This means the columns of A are ... you should know what. If the rank of A is less than 8 (in this case, or less than n in the general case) then the columns of A are not "you know what" and there will be how many solutions?
I hope this is clear. You seem to be using slightly different nomenclature than I'm used to.
If b = the 0 vector, then yes, the system is consistent, but the only solution is the trivial one. If the rank is 8 (in this case, or n in the general case) that will be the only solution. This means the columns of A are ... you should know what. If the rank of A is less than 8 (in this case, or less than n in the general case) then the columns of A are not "you know what" and there will be how many solutions?
I hope this is clear. You seem to be using slightly different nomenclature than I'm used to.