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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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Note that the solution space is not a linear subspace except if $c=d=0$. However, the dimension of the solution space as an affine subspace is indeed $1$, which coincides with the number of free variables.

We can deduce that the first two columns of our original matrix are linearly independent, but that the third lies in their span. That is, the third vector lies in the plane spanned by the first two vectors. If $\alpha=0$ or $\beta=0$, we may further deduce that the last vector is a multiple of one of the first two, but this need not be true in general.

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  • $\begingroup$ Could you explain what do you mean by "the dimension of the solution space as an affine subspace", please? $\endgroup$ – Ninja Mar 6 '17 at 14:50
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    $\begingroup$ @Ninja I mean that the solution space is an affine subspace, and that the dimension of this subspace is $1$. $\endgroup$ – Omnomnomnom Mar 6 '17 at 15:35
  • $\begingroup$ What exactly is an affine subspace? $\endgroup$ – Mary Star Mar 6 '17 at 16:03
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    $\begingroup$ @MaryStar Google it. In a nutshell, though: it's a linear space except that it might be shifted away from the origin. Rather than saying it's closed under linear combinations, we say that the line through any two points in the space is contained in the space. $\endgroup$ – Omnomnomnom Mar 6 '17 at 16:27
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$$x+\alpha z=c$$ is a plane parallel to the axis $y$.

$$y+\beta z=d$$ is a plane parallel to the axis $x$.

$$0=0$$ is the whole space.

The system is the intersection of these three sets, i.e. an oblique straight line (dimension $1$).

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