Prove that if the columns are linearly dependent then $det(A)=0$ Let $A=(a_{ij})\in M_n$ be an arbitrary matrix and let 
$A_1=\begin{pmatrix}
a_{11}\\
a_{21}\\
\vdots\\
a_{n1}\\
\end{pmatrix}$
$A_2=\begin{pmatrix}
a_{12}\\
a_{22}\\
\vdots\\
a_{n2}\\
\end{pmatrix}\ldots$ 
$A_n=\begin{pmatrix}
a_{1n}\\
a_{2n}\\
\vdots\\
a_{nn}\\
\end{pmatrix}\in M_{n1}$ be columns of $A$. Prove that if the set$\{A_1,A_2,...,A_n\} $ is linearly dependent in vector space $M_{n1}$, then $\det A=0$.
I know this already has an answer here but I don't understand OP's solution. 
$\lambda_1 A_1 + \ldots + \lambda_n A_n = 0$ where not all $\lambda_i$ are zero.
Suppose that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = - \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n. \end{align*}
Now what happens after that, with the determinant? 
 A: Since exchanging two columns only switches sign to the determinant, it is not restrictive to assume that the last column is a linear combination of the previous $n-1$ columns:
$$
A_n=\alpha_1A_1+\dots+\alpha_{n-1}A_{n-1}
$$
By multilinearity of the determinant, you have
$$
\det A=
\det\begin{bmatrix} A_1 & \dots & A_{n-1} & 
\sum\limits_{i=1}^{n-1}\alpha_iA_i\end{bmatrix}=
\sum_{i=1}^{n-1}\alpha_i
\det\begin{bmatrix} A_1 & \dots & A_{n-1} & A_i\end{bmatrix}=0
$$
because a matrix with two equal columns has zero determinant again by the above mentioned property above that exchanging two columns changes the sign of the determinant.
With $\begin{bmatrix} v_1 & \dots & v_{n-1} & v_n\end{bmatrix}$ I denote the matrix whose columns are the column vectors $v_1,\dots,v_{n-1},v_n$.
A: I'll give a variant of the proof, hoping you'll understand better.
Suppose there's a non-trivial linear relation between the columns: 
$$\lambda_1A_1+\lambda_2A_2+\dots+\lambda_nA_n=0.$$
Say $\lambda_1\ne 0$. By linearity w.r.t. the 1st column, $\;\det(
\lambda_1A_1,A_2,\dots,A_n)=\lambda_1\det(A_1,A_2,\dots,A_n)$. Also, the determinant is alternating and linear in each column, so
$$ \lambda_1\det A=\det(\lambda_1A_+\lambda_2A_2+\dots+\lambda_nA_n,A_2,\dots,A_n)= \det(0,A_2,\dots,A_n)=0,$$
whence $\det A=0$.
A: The determinant of $A$ is the product of the elements of the main diagonal when $A$ is converted to row echelon form. For a linearly dependent set of columns, when $A$ is converted to row echelon form, there will be a $0$ in the main diagonal of the matrix corresponding to the column of the free variable and hence, the determinant is $0$.
