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    Quote Originally Posted by Klaas van Aarsen View Post
    Ah good.

    Does the solution for $x$ match the given general solution?
    Yes, it does...

    So is this sufficient to say that since it has to hold for every $\lambda$, it has to hold that $a_{12}=a_{22}=0$ and so the matrix $A$ is $A=\begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}$, since this is confirmed by replacing the matrix that we got at $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?

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    #12
    Quote Originally Posted by evinda View Post
    Yes, it does...

    So is this sufficient to say that since it has to hold for every $\lambda$, it has to hold that $a_{12}=a_{22}=0$ and so the matrix $A$ is $A=\begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}$, since this is confirmed by replacing the matrix that we got at $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?
    Yep.

    I'd say 'and this is confirmed...' rather than 'since this is confimed...' though.

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    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.

    I'd say 'and this is confirmed...' rather than 'since this is confimed...' though.
    Nice... Thank you very much!!!

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