How do I solve this Matrice/Vectors question? Give the matrix for the following linear operator: $A \vec{x} = (3,2,6) \times \vec{x}$, where $\vec{x}$ is any arbitrary vector.

Let $\vec{x}=(x_1,x_2,x_3)^T,$ then we know $$(3,2,6)^T\times \vec{x} =(-6 x_2 + 2 x_3, 6 x_1 - 3 x_3, -2 x_1 + 3 x_2)^T$$ So we can see that $$A(1,0,0)^T=(0,6,-2)^T$$ $$A(0,1,0)^T=(-6,0,3)^T$$ $$A(0,0,1)^T=(2,-3,0)^T$$ So these must constitute the columns of $A,$ so $A=\begin{pmatrix} 0 & -6 & 2\\\ 6& 0 & -3\\\ -2 & 3 & 0 \end{pmatrix}.$

(We need to use the transpose since matrices, conventionally, don't act on row vectors).