### Instruction: Find the matrix $\mathbf{M}$ such that
\[\mathbf{M} \mathbf{v} = -5 \mathbf{v}\]for all vectors $\mathbf{v}.$.### Response: I know that a matrix that satisfies this equation for all vectors $\mathbf{v}$ is called a scalar multiple of the identity matrix, because it scales every vector by the same factor.
So I can write $\mathbf{M} = k \mathbf{I}$, where $k$ is some constant and $\mathbf{I}$ is the identity matrix.
To find the value of $k$, I can plug in any nonzero vector $\mathbf{v}$ into the equation and see what happens.
For example, let $\mathbf{v} = \begin{bmatrix}1\\0\end{bmatrix}.$
Then I get \[k \mathbf{I} \begin{bmatrix}1\\0\end{bmatrix} = -5 \begin{bmatrix}1\\0\end{bmatrix}\]
Expanding the matrix multiplication, I get \[k \begin{bmatrix}1\\0\end{bmatrix} = -5 \begin{bmatrix}1\\0\end{bmatrix}\]
This implies that $k = -5$ by equating the corresponding components of the vectors.