I am having a bad time with this matrix and vector situation, and I think the solution is kind a trick in some part of the computation, but I don't know how to find this:
Find 3 vectors (different than Zero), $\vec{x},\vec{y},\vec{b} \in \mathbb{R}^3 $, where $A\vec{x}=\vec{b}$ and $B\vec{y}=\vec{b}$ $$A =\begin{bmatrix} 0 && 1 && 2\\ 1 && 1 && 1\\ 2 && 1 && 0 \end{bmatrix}$$ $$B =\begin{bmatrix} 1 && -1 && 0\\ 0 && 0 && 1\\ 1 && 1 && 0 \end{bmatrix}$$
Basically, what I have done, is multiplying the matrices by a vector $\vec{X} = (X_1, X_2, X_3)$ and the other matrix by a vector $\vec{Y} = (Y_1, Y_2, Y_3)$.
Then, as I am multiplying a matrix $(3 \times 3)$ with a vector $(3 \times 1)$, my result is a new vector $(3 \times 1)$.
But then, I don't know what to do next, can someone help me with a kind advice on this?
I would start by looking for candidates for $\vec b$. For the equation $A\vec x=\vec b$ to have any solution at all, $\vec b$ must lie in $A$’s column space, and likewise it must also lie in $B$’s column space. $B$’s column space is clearly all of $\mathbb R^3$, but the columns of $A$ are linearly dependent (the sum of the outer columns is twice the central column). So, pick any convenient-looking multiple of any column of $A$ (or a convenient-looking linear combination of them) as $\vec b$ and then solve the two equations for $\vec x$ and $\vec y$. For any particular choice of $\vec b$, there’s a degree of freedom in the choice of a corresponding $\vec x$, but if you construct $\vec b$ as I suggest you should be able to write down a value for $\vec x$ without doing any more work. (Hint: interpret the product $A\vec x$ as a linear combination of the columns of $A$.)