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    Am I overthinking this massively? Just can't seem to figure out what to do?
    Help please asap with the "show further" bit, first bit I've done.
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    This doesn't look right: if you set V = \begin{pmatrix}1 \\ 0 \end{pmatrix} you get XV = \begin{pmatrix}4 \\ 5 \end{pmatrix} which doesn't equal xV for any choice of x.

    What would be true is that if it were the case that XV = xV, then it would follow that x has to satisfy x^2-6x-7 = 0.

    Edit: I.e. the end part of the question would need to be worded ",,, and XV = xV then x satisfies the equation..."
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    Do you know how to find eigenvalues and eigenvectors?
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    (Original post by joepiekos)
    Am I overthinking this massively? Just can't seem to figure out what to do?
    Help please asap with the "show further" bit, first bit I've done.
    Write down the product {\bf XV} and equate the entries with the product x{\bf V}

    You should end up with 4a+3b=xa and 5a+2b=xb and now you want to construct a single quadratic in x by substituting in for a or b from one equation into the other.
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    (Original post by DFranklin)
    This doesn't look right: if you set V = \begin{pmatrix}1 \\ 0 \end{pmatrix} you get XV = \begin{pmatrix}4 \\ 5 \end{pmatrix} which doesn't equal xV for any choice of x.

    What would be true is that if it were the case that XV = xV, then it would follow that x has to satisfy x^2-6x-7 = 0.

    Edit: I.e. the end part of the question would need to be worded ",,, and XV = xV then x satisfies the equation..."
    Exactly what I was thinking!
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    (Original post by RDKGames)
    Write down the product {\bf XV} and equate the entries with the product x{\bf V}

    You should end up with 4a+3b=xa and 5a+2b=xb and now you want to construct a single quadratic in x by substituting in for a or b from one equation into the other.
    I guess I was overthinking then Thanks a lot
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    (Original post by B_9710)
    Do you know how to find eigenvalues and eigenvectors?
    No I don't sorry
 
 
 
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