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    #1
    Hey!!

    We have the tableau $\begin{pmatrix}
    \left.\begin{matrix}
    1 & 0 & \alpha \\
    0 & 1 & \beta \\
    0 & 0 & 0
    \end{matrix}\right|\begin{matrix}
    c\\
    d\\
    0
    \end{matrix}
    \end{pmatrix}$

    Since there is a zero-row, we conclude that the column vectors are linearly dependent.

    The number of linearly independent row- and column vectors is the same. And from the tableau we get that there are $2$ linearly independent row- and column vectors. Therefore the dimension of the the vector space spanned by the solumn vectors is $2$. And this is also equal to the rank of the matrix.

    The dimension of the solution space is equal to the numer of free variables, so $1$, right?

    Is everything correct so far?

    What is a geometric interpretation of all these information? Do we get that two column vectors are either a multiple of each other or they are on the same line? Or is there also an other interpretation?

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