Krdytkyu

I wrote out A as a matrix looking like this:
$$begin{bmatrix}a_1&a_2&a_3\a_4&a_5&a_6end{bmatrix}$$
And then using the equation $Atimes[1;1;1]^T=[0;0]^T$ I came to this system of equations:

$$begin{cases}
a_1+a_2+a_3=0\
a_4+a_5+a_6=0
end{cases}
$$

So basically $A$ could be a matrix like this:

$$begin{bmatrix}0&1&-1\-3&7&-4end{bmatrix}$$

or this:
$$begin{bmatrix}-10&6&4\-10&-5&15end{bmatrix}$$
etc...

So I am able to find examples of such $A's$ but I don't know how to find all these $A's$.

I also need to find the dimension and basis for the linear space that these matrices create - the dimension is $6$ but what is the basis?

Here are some hints to get you going in the right direction:

To find a basis, you need to find some examples (you've got those!) that are linearly independent, and that span the subspace. One way to arrange for linear independence is to pick particularly easy examples. In $Bbb R^3$, we choose $pmatrix{1\0\0}, pmatrix{0\1\0}, pmatrix{0\0\1}$ because it's really easy, with all those zeroes, to see that the three are linearly independent.

Can you, first of all, find some elements of your subspace (let's call it $S$) whose entries are mostly zeroes? (All zeroes is of no use: the zero vector is not part of any set of linearly independent vector!)

Second, you need to find the dimension, right? Well, the dimension of the space $M$ of $2 times 3$ matrices is six, and there are some matrices in $M$ that are not in $S$, like
$$
pmatrix{1 & 3 & 1 \ 0 & 1 & 0}.
$$
So the dimension of $S$ must be less than six. Suppose you could find a linearly independent set of vectors in $S$ that had four elements. Then you'd know that the dimension of $S$ was at least four (because the dimension is the size of the largest possible set of linearly independent vectors). Could it be five? Sure ... maybe these one more independent vector to be found. Could it be three? No way.

So what you want to do is look for large sets of linearly independent vectors (i.e., matrices in $S$). Perhaps your work on the "mostly zero" matrices in $S$ will help you here.

Notice that:

$$begin{cases}
a_1+a_2+a_3=0\
a_4+a_5+a_6=0
end{cases}$$

is a system of $2$ equations and $6$ variables. We can introduce $6-2 = 4$ real parameters, $x$, $y$, $z$ and $w$, such that:

$$begin{cases}
a_1 = -x-y\
a_2 = x\
a_3 = y\
a_4 = -z-w\
a_5 = z\
a_6= w
end{cases}.$$

Therefore, all matrices $A$ which satisfy you requirements are in the form:

$$begin{bmatrix}-x-y&x&y\-z-w&z&wend{bmatrix},$$

for given parameters $x, y, z$ and $w$.

I also have a second part to this taks that I don't really understand.

Let $V$ be the space that all of the $A$ matrices create. Now I'm looking for all matrices $B$ that satisfy the equation: $Btimes[1;1;1]^T=[6;6]^T$ and I need to show the set of all $B's$ as a translation of the space $V$ by a six-dimensional vector $b$.

How would this vector $b$ look?

I know that all $B$ matrices are in the form:

$$begin{bmatrix}a&b&6-a-b\c&d&6-c-dend{bmatrix}$$

but I don't think that this gives me anything...

The row vectors of $A$ must be orthogonal to $(1,1,1)^t$. The orthogonal complement of this vector is spanned by $(1,0,-1)^t$ and $(1,-1,0)^t$, for example. Now take as the row vectors of $A$ any two linear combinations of these vectors.

Explicitely, the first row is a linear combination of both vectors, say $x(1,0,-1)^t+y(1,-1,0)^t=
(x+y,-y,-x)^t$ as well as the second row, say
$u(1,0,-1)^t+v(1,-1,0)^t=
(u+v,-u,-v)^t$, similar to the_candyman's answer.

You may write explicitly
$$begin{pmatrix}x&y\ u&vend{pmatrix}
begin{pmatrix}1&0&-1\ 1&-1&0end{pmatrix}$$
for an element of the kernel.

To answer your second part: Let $Tcolon M(2,3)tomathbb R^2$ defined by $T(A):=A(1,1,1)^t$, that is, $T$ is linear and maps a $(2times3)$-matrix to a $2$-dimensional vector, namely the sum of its columns. From this point of view the first task was to find the kernel of $T$. (This was given in my first answer.) Explictely, the kernel of $T$ consists of all matrices for which the sum of the column vectors is $(0,0)^t$.

In your second question you are asked to solve a Linear System, namely $T(A)=(6,6)^t$. As you know, all solutions of a linear system are achieved in adding to a particular solution an element of the kernel. Now a particular solution is any matrix for which the sum of the column vectors is $(6,6)^t$. The general solution is then given by
$$begin{pmatrix}x&y\ u&vend{pmatrix}
begin{pmatrix}1&0&-1\ 1&-1&0end{pmatrix}
+begin{pmatrix}6&0&0\ 6&0&0end{pmatrix}.$$

Does that give you something?