Answer
Because columns of the matrix $B$ are linearly dependent, there are non-trivial constants such that linear combination of those columns is equal to 0 . Using the same constants show that columns of matrix $A B$ make a linear combination equal to 0 which means that they are linearly dependent.
Work Step by Step
Put $B=\left[\begin{array}{llll}b_{1} & b_{2} & \cdots & b_{n}\end{array}\right]$ where $b_{1}, b_{2}, \dots, b_{n}$ are columns of the matrix
$B .$ Agreeing to proposition those columns are linearly dependent, thus there are some constants $c_{1}, c_{2}, \ldots c_{n}$ like $c_{1} b_{1}+c_{2} b_{2}+\dots+c_{n} b_{n}=0$ and $c_{i} \neq 0$ for some $i \in\{1,2, \ldots, n\} .$ The result of matrices $A$ and $B$ is then:
\[
A B=\left[\begin{array}{llll}
A b_{1} & A b_{2} & \cdots & A b_{n}
\end{array}\right]
\]
where the $i^{\text {th }}$ column of the matrix $A B$ is result of the matrix $A$ and the
$i^{\text {th }}$ column of the matrix $B$. Now we have that:
\[
\begin{aligned}
c_{1} A b_{1}+c_{2} A b_{2}+\cdots+c_{n} A b_{n} &=A c_{1} b_{1}+A c_{2} b_{2}+\cdots+A c_{n} b_{n}=\\
&=A\left(c_{1} b_{1}+c_{2} b_{2}+\cdots+c_{n} b_{n}\right)=\\
&=A(0)=0
\end{aligned}
\]
There exist constants $c_{1}, c_{2}, \ldots, c_{n}$ like $c_{1} A b_{1}+c_{2} A b_{2}+\dots+c_{n} A b_{n}=0$
and $c_{i} \neq 0$ for some $i \in\{1,2, \ldots, n\},$ so columns of the matrix $A B$ are linearly dependent.
