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The task I have has 2 parts, I did the first one but now I'm struggling with the second one.

In the first part, I was supposed to find all $2\times3$ matrices $A$ that satisfy the equation: $A\times[1\;1\;1]^T=[0\;0]^T$. I did that, the $A$ matrices are in the form: $$\begin{bmatrix}x&y&-x-y\\z&w&-z-w\end{bmatrix}$$

Then I was supposed to find the basis and dimension for the linear space $V$ that these matrices create, which I also did - dimension is $4$ and the basis is:

$$\{\begin{bmatrix}1&0&-1\\0&0&0\end{bmatrix},\begin{bmatrix}-1&1&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\1&-1&0\end{bmatrix},\begin{bmatrix}0&0&0\\-1&0&1\end{bmatrix}\}$$

Now in the second part, I'm looking for all $B$ matrices that satisfy this equation: $B\times[1\;1\;1]^T=[6\;6]^T$ and I'm supposed to present the set of all the $B's$ as as a translation of the space $V$ by a six-dimensional vector $b$.

How would this vector b look?

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$b$ would look like $\begin{bmatrix}b_1&b_2&b_3\\b_4&b_5&b_6\end{bmatrix}$. Or if you like $b=(b_1,\dots,b_6)$.

You need to find the $b_i$. $b$ should be a particular solution of the second equation.

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  • $\begingroup$ b would look something like this (I did the same thing as in the first part): $$\begin{bmatrix}x&y&6-x-y\\z&w&6-z-w\end{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$\begin{bmatrix}1&2&3\\5&0&1\end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$\alpha\times\begin{bmatrix}1&0&-1\\0&0&0\end{bmatrix}+\beta\times\begin{bmatrix}-1&1&0\\0&0&0\end{bmatrix}+\gamma\times\begin{bmatrix}0&0&0\\1&-1&0\end{bmatrix}+\delta\times\begin{bmatrix}0&0&0\\-1&0&1\end{bmatrix}+\begin{bmatrix}1&2&3\\5&0&1\end{bmatrix}?$$ $\endgroup$ – agromek Nov 24 '18 at 17:44
  • $\begingroup$ @agromek I believe so. That is, yes. $\endgroup$ – Chris Custer Nov 24 '18 at 17:49
  • $\begingroup$ Ok, thank you very much:) $\endgroup$ – agromek Nov 24 '18 at 17:50
  • $\begingroup$ You're welcome. Adding any element of the kernel to a particular solution gives another solution. $\endgroup$ – Chris Custer Nov 24 '18 at 17:53

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