Artin's proof of linearity of determinant in rows of matrix Definition of linearity: Let $A_i$ denote the $i$th row of matrix $A$. Let $A, B, D$ be matrices, all of whose entries are equal except for those in row $k$. Suppose furthermore that $D_k = cA_k + c'B_k$ for scalars $c, c'$. Then $\det D  = c\det A  + c'\det B $
Artin gives a proof by induction. He assumes that this property is true for all $n-1 \times n-1$ matrices and then tries to show that $d_{v1} \det(D_{v1}) = ca_{v1}\det(A_{v1}) + c'b_{v1}\det(B_{v1})$ for every index $v$. He splits the problem into two cases, $v=k$ (which I understand) and $v\not=k$ (which I don't).
Here is his explanation of the second case:
If we let $A'_k, B'_k, D'_k$ denote the vectors obtained from rows $A_k, B_k, D_k$ respectively by dropping the first entry, then $A'_k$ is the row of the minor $A_{v1}$, etc. I understand this concept. However, I don't understand the rest of the proof:
Here $D'_k = cA'_k + c'B'_k$ and by induction on n, $\det D'_{v1}  = c\det(A'_{v1}) + c'\det B_{v1}$. On the other hand, since $v\not=k$, $a_{v1}, b_{v1}, c_{v1}$ are equal. So $d_{v1} \det D_{v1} = ca_{v1}\det A_v1 + c'b_{v1}\det B_{v1}$
Specifically, I don't understand the induction step, and how to get from $\det D'_{v1}  = c\det A'_{v1} + c'\det B_{v1}$ to $d_{v1} \det D_{v1} = ca_{v1}\det A_{v1} + c'b_{v1}\det(B_{v1})$.
 A: I understand that he is calculating a determinant by "expanding" the first column:
$$\left|\begin{matrix}d_{11} & d_{12} & \dots & d_{1n}\\
d_{21} & d_{22} & \dots & d_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
d_{n1} & d_{n2} & \dots & d_{nn}\end{matrix}\right| = d_{11}\left|\begin{matrix}
 d_{22} & \dots & d_{2n}\\
 \vdots & \ddots & \vdots\\
 d_{n2} & \dots & d_{nn}\end{matrix}\right| - d_{21}\left|\begin{matrix}
 d_{12} &d_{13} & \dots & d_{1n}\\
d_{32} & d_{33}
& \dots & d_{3n} \\
\vdots & \vdots & \ddots & \vdots\\
 d_{n2} & d_{n3} & \dots & d_{nn}\end{matrix}\right| + \dots + (-1)^{n+1}d_{n1}\left|\begin{matrix}
 d_{12} &d_{13} & \dots & d_{1n}\\
d_{22} & d_{23}
& \dots & d_{2n} \\
\vdots & \vdots & \ddots & \vdots\\
 d_{(n-1)2} & d_{(n-1)3} & \dots & d_{(n-1)n}\end{matrix}\right|$$
The left hand side is $\det(D)$, and in the right hand side, the $s$-th summand is $\pm d_{s1}$ multiplied by the determinant of the matrix $D$ "without the first column and $s$-th row".
You could write the same expression for the matrices $A$ and $B$. The $k$-th summands of these three expansions are multiplied by the same "minor" (the same $(n-1)\times(n-1)$ determinant, and that is because that determinant is obtained by erasing the only row which is different between $A, B, D$), and we have $d_{k1} = ca_{k1} + c'b_{k1}$, so you can see that the $k$-th summand for $D$ is $c$ times the $k$-th summand for $A$ plus $c'$ times the $k$-th summand for $B$. This is the part you understand.
Now, for $s\neq k$, let's compare the $s$-th summand in the expansion for $D$ with the correspondings summands for $A$ and $B$.
By hypothesis, $d_{s1} = a_{s1} = b_{s1}$, because $s$-th row is equal for the three matrices. So, we must see that the $s$-th minor of $D$ is the required linear combination of the $s$-th minors for $A$ and $B$. Let's denote by $m(A)_s, m(B)_s, m(D)_s$ these matrices, obtained from $A$, $B$ and $D$ by eliminating the first column and $s$-th row.
These three matrices are equal except for one row $j$. The position $j$ of this row is not too important. If $s>k$, then we have taken out a row below the $k$-th one, so $j = k$ retains the same position. If $s<k$, then we have deleted a row above the $k$-th one, so $j = k-1$.
Anyway, these matrices are $(n-1)\times(n-1)$ matrices, equal except for the row $j$, and we have that the $j$-th row of $m(D)_s$ is equal to $c$ times the $j$-th row of $m(A)_s$ plus $c'$ times the $j$-th row of $m(B)_s$. By induction hypothesis, the theorem is true for these matrices, that is:
$$\det(m(D)_s) = c\det(m(A)_s) + c'\det(m(B)_s)$$
This proves the assertion, because, as we saw, in each expansion these minors are multiplied by the same $a_{s1} = b_{s1} = d_{s1}$.
A: Here is an answer (as opposed to my comment response above).
It is important to keep in mind that for $v \neq k$ that $D_{vj} = A_{vj} = B_{vj}$ and similarly $d_{v1} = a_{v1} = b_{v1}$.
Furthermore, for $v=k$, we have $d_{vj} = ca_{vj} + c' b_{vj}$.
The cofactor expansion gives $\det D = \sum_{v=1}^n (-1)^{v+1} d_{v1}\det D_{v1}$, so showing that
$d_{v1}\det D_{v1} = c a_{v1}\det A_{v1} + c' b_{v1}\det B_{v1}$ is certainly sufficient to show that
$\det D = c \det A + c' \det B$.
Case $v=k$:
\begin{eqnarray}
d_{v1}\det D_{v1} &=& (ca_{v1} + c' b_{v1}) \det D_{v1}\\
&=& ca_{v1} \det D_{v1}+ c' b_{v1}\det D_{v1} \\
&=& ca_{v1} \det A_{v1}+ c' b_{v1}\det B_{v1}
\end{eqnarray}
Case $v \ne k$:
We have $d_{v1} = a_{v1} = b_{v1}$, so focus on $\det D_{v1}$. Remember that the $k$ row has elements $d_{kj} = ca_{kj} + c' b_{kj}$ and all other elements of $D_{v1}$ equal the corresponding elements of both $A_{v1}$ and $B_{v1}$.
In particular, the $(n-1) \times (n-1)$ matrix $D_{v1}$ has the form of being equal everywhere except for Row $k$ which is the sum of elements as above. Hence the induction step tells us that
$\det D_{v1} = c \det A_{v1} + c' \det B_{v1}$ and so
\begin{eqnarray}
d_{v1}\det D_{v1} &=& d_{v1} (c \det A_{v1}+ c' \det B_{v1}) \\
&=& d_{v1} c \det A_{v1}+ d_{v1} c' \det B_{v1} \\
&=& a_{v1} c \det A_{v1}+ b_{v1} c' \det B_{v1}
\end{eqnarray}
as required.
