# Determinant proof

Let $A\in M_n(\mathbb C)$ and $\alpha \in \mathbb C$. If $B$ is the matrix obtained by multiplying a single row of $A$ by $\alpha$, then det$(B)=$ $\alpha$ det$(A)$.

I'm trying to understand and use this definition of determinant,

$$det(A)= \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n a_{i \sigma(i)}$$

To try to understand better, I considered a 2 $\times$2 matrix $$\begin{pmatrix}a_{11}&a_{21}\\a_{21}&a_{22}\\ \end{pmatrix}$$ But how will it work? If I let $B$ be any row in $A$ multiplied by $\alpha$, I get a 2 $\times$1 matrix, and I can't find the determinant of that since it's not a square matrix...

• You are taking $A$, and then multiplying a row by $\alpha$ and leaving the other rows intact. – Henry Swanson Dec 2 '13 at 7:36
• @HenrySwanson Ohh okay, I get it now. Thanks! – Alti Dec 2 '13 at 8:05
• This is not hard to see, so I'll only describe how it goes: observe that for each term of the sum we have a product on which one, and only one, element of each row appears, so we have that the term $\alpha$, that is multiplying all the terms of one row, will appear one time for each term of the sum, so you can put it on evidence and it goes out of the determinant as expected. – user119459 Aug 15 '14 at 5:31