### Instruction: Find the matrix $\mathbf{M}$ such that
\[\mathbf{M} \mathbf{v} = \begin{pmatrix} 5 \\ 2 \\ -8 \end{pmatrix} \times \mathbf{v}\]for all vectors $\mathbf{v}.$.### Response: To find the matrix $\mathbf{M}$, I need to figure out what it does to the basis vectors $\mathbf{i} = (1,0,0), \mathbf{j} = (0,1,0), \mathbf{k} = (0,0,1)$ of $\mathbb{R}^3$.
By plugging in $\mathbf{i}$, I get
\[\mathbf{M} \mathbf{i} = \begin{pmatrix} 5 \\ 2 \\ -8 \end{pmatrix} \times \mathbf{i} = \begin{pmatrix} 0 \\ -8 \\ -2 \end{pmatrix}\]
Similarly, by plugging in $\mathbf{j}$ and $\mathbf{k}$, I get
\[\mathbf{M} \mathbf{j} = \begin{pmatrix} 5 \\ 2 \\ -8 \end{pmatrix} \times \mathbf{j} = \begin{pmatrix} 8 \\ 0 \\ 5 \end{pmatrix}, \quad \mathbf{M} \mathbf{k} = \begin{pmatrix} 5 \\ 2 \\ -8 \end{pmatrix} \times \mathbf{k} = \begin{pmatrix} 2 \\ -5 \\ 0 \end{pmatrix}\]
Now, I recall that the columns of a matrix are the images of the basis vectors under the matrix multiplication.