Linearity property of Determinant function:
For an $n\times n$ matrix $A$, we can consider det$A$ as a function of the $n$ column vectors in $A$. We will show that if all columns except one are held fixed, then det$A$ is a linear function of that one (vector) variable.
Suppose that the $j$th column of $A$ is allowed to vary, and write
$A = [\textbf{a}_{1}\quad \cdots\quad \textbf{a}_{j-1}\quad \textbf{x} \quad \textbf{a}_{j+1}\quad \cdots \quad \textbf{a}_{n}]$
Define a transformation $T$ from $\mathbb{R}^{n}$ to $\mathbb{R}$ by
$T(\textbf{x}) = \text{det}[\textbf{a}_{1}\quad \cdots\quad \textbf{a}_{j-1}\quad \textbf{x} \quad \textbf{a}_{j+1}\quad \cdots \quad \textbf{a}_{n}]$
Then,
$T(c\textbf{x}) = c(T\textbf{x})\quad$ for all scalars $c$ and all $\textbf{x}$ in $\mathbb{R}^{n}$
$T(\textbf{u}+\textbf{v}) = T(\textbf{u}) + T(\textbf{v}) \quad$ for all $\textbf{u}$, $\textbf{v}$ in $\mathbb{R}^{n}$
Source: Linear Algebra and Its Applications by David C. Lay
Q: Use the linearity of the determinant to explain why: if a multiple of one row of $A$ is added to another row to produce a matrix $B$, then $|B| = |A|$.
Not sure if it has to do with the fact that if one row of $A$ is multiplied by $k$ to produce $B$,
then det$B$ = $k\times \text{det}A$?
The property key to understanding this is the fact that the determinant of a Matrix with two identical rows is $0$:
$T(a(i)) = \det{[a(1)\ldots a(j-1) \quad a(i) \quad a(j+1)\ldots a(n)]} = 0$ for $i\neq j $
This can be proved by permuting the free column, that we have set to $a(i)$, with the fixed $i^{th}$ row. We obtain a new $T'(a(i))= -T(a(i))$, but both determinants are equal so the only possibility is $T(a(i))=0$.
Then:
$T(k a(i)+ x) = \\ \det{[a(1) ..a(j-1) \quad ka(i) \quad a(j+1) .. a(n)]} + \det{[a(1) .. a(j-1) \quad x \quad a(j+1) .. a(n)]} = \\ \det{[a(1) .. a(j-1) \quad x \quad a(j+1) .. a(n)]} $