Proving determinant is zero 
Let $n\ge 2$ and $A=[\overset{\to}{a_1},\overset{\to}{a_2},\ldots,\overset{\to}{a_n}]$ an $n\times n$ matrix such that there exists $i\neq j$ such that $$\overset{\to}{a_j}=k\overset{\to}{a_i}$$
where $k\neq 0$. Show that $\det(A)=0$.

How would I approach this proof or solve it?

If $a_i=ka_j$ for some $k\neq 0$ then $a_i$ and $a_j$ are linearly dependent. This imply that the rank of the matrix is less than $n$, thus if $A$ represent a linear operator it cannot be injective, hence it is not invertible and $\det(A)=0$.
P.S.: Im not sure that this answer will be useful for you, I take a little "roundabout" when I talk about operator representation. In any case the above assume that you knows that if $A$ is not invertible then it determinant is zero.
