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Math Help - Trouble with Linear Alg ideas in my Diffy Q Class

  1. #1
    No one in Particular VonNemo19's Avatar
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    Trouble with Linear Alg ideas in my Diffy Q Class

    Ji everyone. I'm taking differential eqns right now and I never took linear algebra and my teacher is assuming that all of the students have had it before, so I have just a quick question:

    Why is it the case that if the determinant of a set of n vectors is 0 then the set is linearly dependent. I understand how to apply this concept, I'm just having trouble with the "why".
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Ji everyone. I'm taking differential eqns right now and I never took linear algebra and my teacher is assuming that all of the students have had it before, so I have just a quick question:

    Why is it the case that if the determinant of a set of n vectors is 0 then the set is linearly dependent. I understand how to apply this concept, I'm just having trouble with the "why".
    Ok, by determinant of a set of n vectors I assume you mean the determinant of the matrix you get when you use these vectors as rows or columns. in that case, here is the explanation. I'll make it concise, so ask questions if something is not clear.

    ...

    actually, it after typing out huge chunks of code, it became apparent to me that MHF is currently having issues with LaTeX. So I'll just tell you how to see it for yourself with hints and directions.

    What does it mean for a set of vectors to be linearly dependent? (Hint: your answer should involve an equation and a set of solutions for that equation)

    Set this equation up for your set of vectors. Create a system of equations from it, and augment the system into matrices. The matrix to the far left should be the matrix of your vectors.

    Now, assume that the determinant of this matrix is zero. This is equivalent to several things. (1) it means the matrix does NOT reduce to the identity, (2) it is not invertible and (3), perhaps the most important here, is that the system won't have a unique solution. It will have many. Now consider what this means based on the first question I asked you
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Jhevon View Post
    Ok, by determinant of a set of n vectors I assume you mean the determinant of the matrix you get when you use these vectors as rows or columns. in that case, here is the explanation. I'll make it concise, so ask questions if something is not clear.

    ...

    actually, it after typing out huge chunks of code, it became apparent to me that MHF is currently having issues with LaTeX. So I'll just tell you how to see it for yourself with hints and directions.

    What does it mean for a set of vectors to be linearly dependent? (Hint: your answer should involve an equation and a set of solutions for that equation)

    Set this equation up for your set of vectors. Create a system of equations from it, and augment the system into matrices. The matrix to the far left should be the matrix of your vectors.

    Now, assume that the determinant of this matrix is zero. This is equivalent to several things. (1) it means the matrix does NOT reduce to the identity, (2) it is not invertible and (3), perhaps the most important here, is that the system won't have a unique solution. It will have many. Now consider what this means based on the first question I asked you
    If you have saved your chunk of code you could use www.quicklatex.com or Online LaTeX Equation Editor - create, integrate and download

    I believe the latter allows you to save it as a jpeg for subsequent attachment.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bugatti79 View Post
    If you have saved your chunk of code you could use www.quicklatex.com or Online LaTeX Equation Editor - create, integrate and download

    I believe the latter allows you to save it as a jpeg for subsequent attachment.
    That assumes we can upload attachments. (That one may have been fixed, I haven't checked.) I've been using the method in post 8 of this thread.

    -Dan
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