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Math Help - n by n matrix

  1. #1
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    n by n matrix

    can someone please give me an example of an n x n matrix, where the last column is the sum of the previous n-1 columns?
    i have to prove whether it is invertible or not

    thanks
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  2. #2
    Senior Member roninpro's Avatar
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    Is there something preventing you from doing this on your own?
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  3. #3
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    i'm not sure whether the example i made up was correct

    a11 a12 a13 | (a13+ a23+ a33)
    a21 a22 a23 | (a12 + a22 +a32)
    a31 a32 a33 | (a11 + a21 +a31)

    is this correct?
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  4. #4
    Senior Member roninpro's Avatar
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    That's not an n\times n matrix, though.
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  5. #5
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    woops. sorry i didn't see that

    a11 a12 (a12:a32)
    a21 a22 (a11:a31)
    a31 a32 (...)

    i get stuck when it comes to a33.
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  6. #6
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    Quote Originally Posted by Kbotz View Post
    i'm not sure whether the example i made up was correct

    a11 a12 a13 | (a13+ a23+ a33)
    a21 a22 a23 | (a12 + a22 +a32)
    a31 a32 a33 | (a11 + a21 +a31)

    is this correct?
    i think you should add an extra row
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  7. #7
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    Quote Originally Posted by Kbotz View Post
    can someone please give me an example of an n x n matrix, where the last column is the sum of the previous n-1 columns?
    i have to prove whether it is invertible or not

    thanks
    To disprove something, all you need is an example. Since there aren't many stipulations on this matrix, I can think of many singular nxn matrices that meet your criteria.
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  8. #8
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    Quote Originally Posted by Dgphru View Post
    i think you should add an extra row
    Adding an extra row makes the matrix 4x3 which isn't nxn
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  9. #9
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    Assume this is nxn

    \begin{bmatrix}<br />
a & a & b & \dots & \sum_{x=1}^{n}a_{1x}\\ <br />
a & a & c &  & \\ <br />
a & a & d &  & \vdots\\ <br />
a & a & e & \ddots & \\ <br />
a & a & f &  & \sum_{x=1}^{n}a_{nx}<br />
\end{bmatrix}

    This matrix isn't invertible since column 1 and 2 are lin. dep.
    Last edited by dwsmith; May 22nd 2010 at 08:24 PM.
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    Assume this is nxn

    \begin{bmatrix}<br />
a & a & b & \dots & \sum_{x=1}^{n}a_{1x}\\ <br />
a & a & c &  & \\ <br />
a & a & d &  & \vdots\\ <br />
a & a & e & \ddots & \\ <br />
a & a & f &  & \sum_{y=1}^{n}\sum_{x=1}^{n}a_{yx}<br />
\end{bmatrix}

    This matrix isn't invertible since column 1 and 2 are lin. dep.

    are b,c,d...f the sum of the previous columns?
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  11. #11
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    If they have to be, then yes. By last column, do you mean every column needs to be the sum? I thought you mean column n is the sum of all the columns.

    It doesn't change anything then because column 2 is the sum of column 1 then; thus, column 1 and 2 are lin. ind.
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  12. #12
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    sorry, my initial question was a little vague.

    Let B be an n by n matrix. suppose its last column is the sum of the previous n-1 columns. is it possible for B to be invertible? why or why not?
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  13. #13
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    i found it quite difficult to provide an example since for a n by n matrix, the right hand bottom corner of the matrix (a(mn)) will not have a value.
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  14. #14
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    Quote Originally Posted by Kbotz View Post
    sorry, my initial question was a little vague.

    Let B be an n by n matrix. suppose its last column is the sum of the previous n-1 columns. is it possible for B to be invertible? why or why not?
    What do you mean the last column is the sum of previous?

    Does that mean the nth column is the sum or does that mean column two is the sum of column 1, column 3 is the sum of column 1 and 2....? If the second is the case, you can make the matrix columns the Fibonacci numbers.
    1, 1, 2, 3, 5, 8, ....
    *note v_1 means a nx1 column vector of the nxn matrix
    If you make v_1 a column vector of all 1s, then v_2= v_1, v_3= 2v_1, v_4= 5v_1,....
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  15. #15
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    i'm hoping it's the second case, but i'm not too sure since it says n-1.
    the question itself is quite vague.
    sorry for the trouble
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