# Matrix and vectors (Find 3 Vectors)

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?

• This is easy. You can take $x=y=b=(0,0,0)$. – Dietrich Burde Apr 23 at 15:13
• Thank you @DietrichBurde, I forgot to mention that should be different than Zero. – jguzmaje Apr 23 at 15:38
• The only thing that I can think of is to let $$\vec{x} = \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$$ Do the same for $\vec{y}$. Since $A\vec{x} = \vec{b}$ and $B\vec{y} = \vec{b}$, we have that $A\vec{x} = B\vec{y}$. Doing the expansion will yield three equations with six unknowns. Not solvable. The only other thing I could think of would be to find $A^{-1}$ and $B^{-1}$ because you have equations $\vec{x} = A^{-1}B\vec{y}$ and $\vec{y} = B^{-1}A\vec{x}$. This may just give you the same three equations... – JacobCheverie Apr 23 at 15:41
• No “tricks” necessary, but a solid understanding of the properties of systems of linear equations helps make this much easier. – amd Apr 23 at 19:19

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 $$A$$.)