
MHB Master
#1
March 6th, 2017,
07:33
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 zerorow, 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?

March 6th, 2017 07:33
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