# Linear Algerba - translation of space $V$ by a six-dimensional vector $b$

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?

$$b$$ 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.
• 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: $$\alpha\times\begin{bmatrix}1&0&-1\\0&0&0\end{bmatrix}+\beta\times\begin{bmatrix}-1&1&0\\0&0&0\end{bmatrix}+\gamma\times\begin{bmatrix}0&0&0\\1&-1&0\end{bmatrix}+\delta\times\begin{bmatrix}0&0&0\\-1&0&1\end{bmatrix}+\begin{bmatrix}1&2&3\\5&0&1\end{bmatrix}?$$ – agromek Nov 24 '18 at 17:44