Okay, so I am in a linear algebra course and we are going though the derivation of the determinant function for matrices. I am struggling with some of the properties of determinant functions i.e. functions which are n-linear, alternating, and give a 1 for the identity matrix. The question I am currently stuck on is as follows.
Let $K$ be a commutative ring with identity and $D$ an $n$-linear function on $n\times n$ matrices over $K$. Show that $D(B)=D(A)$, if $B$ is obtained from $A$ by adding a scalar multiple of one row of $A$ to another.
What I have so far is as follows.
Let $A$ be an $n\times n$ matrix. Let $A=(\alpha_1, \cdots,\alpha_i,\cdots,\alpha_j,\cdots, \alpha_n)$ where $a_k$, $k=1,...,n$ denote the rows of $A$. Let $B=(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_n)$ for $a\in K$. That is let $B$ be the matrix formed from $A$ where the $i^{th}$ row of $A$ is replaced by $a$ times the $i^{th}$ of $A$ row plus the $j^{th}$ row of $A$. Then
$D(B)=D(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_n)=aD(\alpha_1,\cdots,\alpha_i,\cdots,\alpha_j,\cdots,\alpha_n)+D(\alpha_1,\cdots,\alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)\text{($n$-linearity) }=aD(A)+0=aD(A).$
Therefore $D(B)=aD(A)$
I am clearly missing something obvious because the $a$ should not be there according to the question but for the life of me I can't see what is going on. I am assuming it is something obvious and easy but I have been stuck for far too long. Thanks.
Your proof seems right to me. But, you must define $B$ as
$$B=(\alpha_1,\cdots,\alpha_i+a\cdot \alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)$$
because $B$ is obtanined by adding $a$ times the row $j$ of $A$ to the row $i$ of $A$. In this case you will get $D(A)=D(B)$ since:
$$D(B)=D(\alpha_1,\cdots,\alpha_i+a\cdot \alpha_j,\cdots,\alpha_j,\cdots,\alpha_n) =D(\alpha_1,\cdots,\alpha_i,\cdots,\alpha_j,\cdots,\alpha_n)+a\cdot D(\alpha_1,\cdots,\alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)=D(B)+a\cdot 0=D(B).$$