Show that $\det(cA)=c^n\det(A)$ $A$ is a square matrix of order $n$. Show that $\det(cA)=c^n\det(A)$, $c\in\mathbb{R}$.
I don't know where to begin. Can anyone offer some guidance?
 A: $\det$ is a multilinear function of the columns (or rows) of a matrix and so
$\det(cA)=\det(cA_1, cA_2, \dots, cA_n)=c\det(A_1,cA_2, \dots, cA_n) = \cdots = c^n \det(A_1,A_2, \dots, A_n) \\= c^n \det(A)$
A: We have $cA = (cI)A$, and the determinant of a product is the product of the determinants.
An alternative route is to talk about $cI$ being a product of $n$ elementary matrices, each altering the determinant by a factor of $c$.
A: $$Ae_1 \wedge \cdots \wedge Ae_n = \det(A)e_1 \wedge \cdots \wedge e_n$$
defines $\det(A)$.
Now let's look at $\det(cA)$:
$$\begin{align}\det(cA)e_1 \wedge \cdots \wedge e_n &= cAe_1 \wedge \cdots \wedge cAe_n \\ &= c^n Ae_1 \wedge \cdots \wedge Ae_n \\ &= c^n \det(A) e_1 \wedge \cdots \wedge e_n \end{align}$$
Which implies $\det(cA) = c^n\det(A)$.$\ \ \ \ \square$
A: You should know that if $B$ is obtained by multiplying a row of $A$ by a constant $c$, then $\det B = c\det A$.
But $cA$ is just what you get by multiplying each of the $n$ rows of $A$ by $c$, so $\det cA = c^n\det A$, right?
