Answer
See below
Work Step by Step
a) Given $A=\begin{bmatrix}
a_{11} & a_{12} \\a_{21} & a_{22}
\end{bmatrix}$ and $c$ is a constant.
Obtain $\det(cA)=\begin{vmatrix}
a_{11} & a_{12} \\a_{21} & a_{22}
\end{vmatrix}\\=ca_{11}ca_{22}-ca_{12}ca_{21}\\
=c^2a_{11}a_{22}-c^2a_{12}a_{21}\\
=c^2(a_{11}a_{22}-a_{12}a_{21})\\
=c^2\det(A)$
b) Given $A$ is an $n × n$ matrix and $c$ is a constant
Let $B=cA\\
\rightarrow det(B)=\sum \sigma(p_1,p_2,...,p_n)b_{1p_1}b_{2p_2}...b_{np_n}$
Since $b_{ij}=ca_{ij} \forall 1\leq i \leq n, 1\leq j\leq n$
we have $det(B)=\sum \sigma(p_1,p_2,...p_n)b_{1p_1}b_{2p_2}...b_{np_n}\\
=\sum \sigma(p_1,p_2,...p_n)ca_{1p_1}ca_{2p_2}...ca_{np_n}\\
=\sum \sigma(p_1,p_2,...p_n)c^na_{1p_1}a_{2p_2}...a_{np_n}\\
=c^n\sum \sigma(p_1,p_2,...p_n)a_{1p_1}a_{2p_2}...a_{np_n}\\
=c^n\det (A)$
