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chrisplusian

  • one year ago

Linear algebra question.... see attachment (part d)

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  1. chrisplusian
    • one year ago
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  2. chrisplusian
    • one year ago
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    I did the reduced row echelon form and it gives that all variables are equal to zero. Is it reasonable to say that if a system has a solution, and its augmented matrix is one of the form that the augmented part is 0, then the solution will have all variables equal to 0?

  3. chrisplusian
    • one year ago
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    I know this is called the "trivial solution" but is there a case of an augmented matrix with the augmented part as zeros, where the solution will not be the trivial one?

  4. jango_IN_DTOWN
    • one year ago
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    hiii

  5. chrisplusian
    • one year ago
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    Hello

  6. jango_IN_DTOWN
    • one year ago
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    i guess you know how to calculate the rank

  7. jango_IN_DTOWN
    • one year ago
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    when the system is homogeneous, the rank of coefficient matrix and the augmented matrix are equal.. so we may only perform operations on the coefficient matrix

  8. chrisplusian
    • one year ago
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    I don't know what you mean by "rank"

  9. jango_IN_DTOWN
    • one year ago
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    calculate the rank of the coefficient matrix by converting it into row echelon form

  10. jango_IN_DTOWN
    • one year ago
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    You know row echelon form??

  11. jango_IN_DTOWN
    • one year ago
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    @chrisplusian

  12. amistre64
    • one year ago
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    the solution to the set is the origin ...

  13. amistre64
    • one year ago
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    3 planes can meet in a number of different ways for a graphical representation. in this case they make a tepee, whose apes is 0,0,0

  14. amistre64
    • one year ago
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    apex ... thats what i get for trying to be lyrical

  15. chrisplusian
    • one year ago
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    Sorry I lost internet connection @jango_IN_DTOWN. I will show you what I did....

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  16. chrisplusian
    • one year ago
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    @amistre64 is there a case where you can have the augmented matrix, with the "augmented portion" (<---- not sure if this is correct terminology) equals zero, but the solution to the system is non-zero?

  17. jango_IN_DTOWN
    • one year ago
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    This is right..

  18. phi
    • one year ago
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    if the matrix is full-rank (its determinant is not zero) , then only the trivial solution will give all zeros. on the other hand if the matrix is not full rank, yes you can have non-trivial solutions

  19. chrisplusian
    • one year ago
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    I have no idea what "rank" is.

  20. chrisplusian
    • one year ago
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    That is not a topic we have covered yet

  21. phi
    • one year ago
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    if you think of each row in the matrix as an equation, then "rank" is the number of "independent" equations if you have not learned this, you eventually will.

  22. chrisplusian
    • one year ago
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    My question was.... if there is an augmented matrix with the augmented portion (the part to the right of the line) that is zero can there be another solution other than the origin? Like can there be a numerical solution? Or a dependant solution? I think I understand geometrically what the ideas are. If I end up with a numerical solution then there is a place that in R^3 the planes intersect and the solution is that point. If I end up with a consistent solution that is dependent than I have a line that at which all three planes intersect, and if I end up with an inconsistent system that implies that the three plans have no mutual point/s of intersection. What I can't decide is if having an augmented matrix with the augmented portion as all zeros actually tells me anything?

  23. chrisplusian
    • one year ago
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    @phi we just discussed the idea of "lead variables" and "free variables", could this somehow be the same idea as "rank"?

  24. jango_IN_DTOWN
    • one year ago
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    @chrisplusian you know minors and cofactors of a matrix right? Then it will be very easy to define rank

  25. chrisplusian
    • one year ago
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    I understand minors and cofactors :) That was what we discussed in our last class meeting.

  26. jango_IN_DTOWN
    • one year ago
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    ok they will define rank in the near classes I am sure.. Ok rank is "the greatest potisive integer r such that that matrix A has at least one non zero minor of order r".

  27. chrisplusian
    • one year ago
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    Ok so maybe I am jumping ahead?

  28. amistre64
    • one year ago
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    if you know of spans, then the matrix here spans R^3, the vectors are independant. also, each equation defines a plane in R^3 such that they all meet at one point. only one point is common to all 3 planes. |dw:1444587665421:dw|

  29. chrisplusian
    • one year ago
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    Are there any things I should know about an augmented matrix with the augmented part as all zeros, that does not require the concept of "rank"?

  30. chrisplusian
    • one year ago
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    @amistre64 what do you mean by spans?

  31. phi
    • one year ago
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    if you have free variables, then you will not have full rank if the number of lead variables is equal to the number of rows (equations) then you have full rank. (rank is the number of lead variables)

  32. amistre64
    • one year ago
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    a span is a set of vectors that allows us to reach all the points in a given space. for R^2 (1,0) and (0,1) are the standard vectors that allow us to span the enirety of a plane. a basis, is a spanning set of vectors that are independant (none can be defined as a linear combination of the others)

  33. phi
    • one year ago
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    You could take a peek at the later chapters in your book

  34. amistre64
    • one year ago
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    if Ax=0 only has the trivial solution .. then A forms a basis, i beleive

  35. chrisplusian
    • one year ago
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    @phi my professor did not elaborate on the free variables, or lead variables yet. It was just a side note that when you are solving a system, any row that has a non-zero value after you have put it in reduced row echelon form, is considered to have a lead variable. If a column does not have a lead variable then it is a free variable. If you have a dependent system, solve for all the free variables and then write it in vector form.

  36. phi
    • one year ago
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    yes, and if you have a dependent system, then expect there to be non-trivial solutions to Ax=0

  37. phi
    • one year ago
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    But I am sure you will see soon in your course.

  38. chrisplusian
    • one year ago
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    So to summarize if you have the augmented portion as zeros, then you will either have the trivial solution or a dependent system?

  39. phi
    • one year ago
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    yes there are other implications: determinant of the dep. sys is 0 inverse of A does not exist

  40. chrisplusian
    • one year ago
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    As far as a numerical solution, a single point, if the augmented portion is all zeros (before the reduced row echelon form) then can you conclude that there will be no single point other than the origin that would be a solution to the system?

  41. amistre64
    • one year ago
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    no

  42. amistre64
    • one year ago
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    solution to 3 planes are either a point, a line, a plane, or nothing

  43. phi
    • one year ago
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    you can use row reduction to help decide that. Otherwise, only God knows

  44. amistre64
    • one year ago
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    3 parallel planes have no common points 3 coincident planes are common at every point etc

  45. chrisplusian
    • one year ago
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    Thank you all :)

  46. amistre64
    • one year ago
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    spose our setup defines a line in space that does not pass thru the origin. like 3x + 4y -2 = 0 x=0 and y=0 is not a solution. if i understand your question correctly

  47. chrisplusian
    • one year ago
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    No what I was asking about was in reference to an entire system.

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