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?