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anonymous

  • 5 years ago

determine if the vector b is in teh span of the columns of the matrix A A= 1 2 3 4 5 6 7 8 9 b= 10 11 12

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  1. shadowfiend
    • 5 years ago
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    I *believe* the way to do this is to see if there is a multiple of each of the column vectors of \(\bf{A}\) that will produce \(\vec{b}\).

  2. shadowfiend
    • 5 years ago
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    But don't quote me on that!

  3. anonymous
    • 5 years ago
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    do you row reduce it?

  4. shadowfiend
    • 5 years ago
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    Hm, no a multiple of the sums. So I believe you have to break down the columns into three vectors: \[\begin{align} \bf{A} = \left[\begin{matrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{matrix}\right] \end{align}\] You turn this into three vectors: \[\begin{align} \vec{a_1} = \left[\begin{matrix}1\\4\\7\end{matrix}\right] & \vec{a_2} = \left[\begin{matrix}2\\5\\8\end{matrix}\right] & \vec{a_3} = \left[\begin{matrix}3\\6\\9\end{matrix}\right] \end{align}\]

  5. shadowfiend
    • 5 years ago
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    Then, if you can find an \(x, y, z\) such that: \[x\vec{a_1} + y\vec{a_2} + z\vec{a_3} = \vec{b}\] Then I believe \(\vec{b}\) is in the span of the columns of \(\bf{A}\).

  6. shadowfiend
    • 5 years ago
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    You end up with a system of 3 equations, one for each of the rows multiplied by x/y/z, and you can solve to see if there is a solution for x/y/z.

  7. anonymous
    • 5 years ago
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    then do u do the agumented of row reduced?

  8. shadowfiend
    • 5 years ago
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    I have no idea! Hehe. Like I said, I don't remember *too* much about this, and I've posted more or less everything I do remember. Sorry :/

  9. anonymous
    • 5 years ago
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    okay thank you :)

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