Verifying that the given map defines a Lie algebra

I am given a matrix $$A =(a,b;c,d)$$ in $$GL(2,\mathbb{C})$$ and a real algebra say, $$V$$ with basis $$X,Y,Z$$ such that $$[X,Y]=0, [X,Z]=aX+bY, [Y,Z]=cX+dY$$ I have to show that $$V$$ is a real Lie Algebra.

My attempt: it's vector space over $$\mathbb{R}$$ (duh!) I think we first need to find $$[X,X], [Y,Y], [Z,Z]$$ which I don't know how...and then use it to show bilinearity. Similarly, first somehow find $$[Y,X], [Z,Y], [Z,X]$$ to verify anti symmetry.

Assuming both bilinearity and anti symmetry, it's sufficient to verify Jacobi Identity for elements $$\alpha, \beta, \gamma$$ where $$\alpha, \beta, \gamma \in {X,Y,Z}$$.

Consider the three cases when each of these three elements are different from one another, all are the same,two of them are same and one different from the other two. Now, using anti symmetry and above computed values of lie bracket helps us verify the Jacobi Identity.

But as trivial as it seems, I don't seem to have a clue about how to show bilinearity and compute lie brackets \$[X,X],... I'd appreciate any hint/s. Please do not post a solution. Thank you very much!

The problem, as stated, makes no sense. But I suspect the the idea is to prove that there is one and only one Lie algebra structure that we can defined on a real vector space $$V$$ with a basis $$\{X,Y,Z\}$$ for which those relations hold. Of course, then we shall have to have:
• $$[X,X]=0$$;
• $$[Y,Y]=0$$;
• $$[Z,Z]=0$$;
• $$[Y,X]=-[X,Y]=0$$;
• $$[Z,X]=-[X,Z]=-aX-bY$$;
• $$[Z,Y]=-[Y,Z]=-cX-dY$$.
So, if $$\alpha,\alpha',\beta,\beta',\gamma,\gamma'\in\mathbb R$$, then\begin{align}[\alpha X+\beta Y+\gamma Z,\alpha'X+\beta'Y+\gamma'Z]&=\alpha\alpha'[X,X]+\alpha\beta'[X,Y]+\cdots+\gamma\gamma'[Z,Z]\\&=(\alpha\gamma'-\alpha'\gamma)[X,Z]+(\beta\gamma'-\beta'\gamma)[Y,Z]\\&=(\alpha\gamma'-\alpha'\gamma)(aX+bY)+(\beta\gamma'-\beta'\gamma)(cX+dY).\end{align}
Therefore, the idea is to prove that $$V$$ endowed with this product is indeed a Lie algebra.