Answer
$det (kA)=k^2det A$
Work Step by Step
Calculate $kA$:
$$\begin{align*}
kA&=k\begin{bmatrix}2&-1\\1&2\end{bmatrix}=\begin{bmatrix}2k&-k\\k&2k\end{bmatrix}.
\end{align*}$$
Calculate $det A$ and $det (kA)$:
$$\begin{align*}
det A&=2\cdot 2-1\cdot (-1)=5\\
det (kA)&=2k\cdot (2k)-k\cdot(-k)=5k^2.
\end{align*}$$
We notice that $det (kA)=k^2det A$.
