Source : Henrik Bachmann's 2020 Linear Algebra Course ( Nagoya University)
Page $2$ of the following notes : https://www.henrikbachmann.com/uploads/7/7/6/3/77634444/la2_2020_lecturenotes_5.pdf
The theorem this question is dealing with is related to the question of the effect of row operations on the determinant of a matrix. More precisely, with the effect of the operation consisting in multiplying the row of a matrx with a scalar $\lambda$.
In order to show that , by performing this multiplication, the determinant of the ( resuling) matrix is also multiplied by $\lambda$, it is required ( in Bachmann's approach) to prove that the following function $F_{A, l}$ is linear :
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$F_{A, l} : R^n \rightarrow R$
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$1\leq l \leq n$
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to every element $x\in R^n$ , $F_{A, l}$ associates the determinant of the matrix obtained from matrix $A\in R^{n\times n}$ by substituting vector $x\in R^n$ for the l-th row of matrix $A$.
I think I can show that this function satisfies the first condition of linearity , namely that $F_{A, l} (x+y) = F_{A, l}(x) + F_{A, l}(y)$.
But I cannot see how the second conditon is satisfied , namely how $F_{A, l}(\lambda x) = \lambda \times F_{A, l}(x)$ .
Sorry for not being able to show my work, I'm struggling to understand the notes in order to see the "big picture" of Linear Algebra, but I'm not familiar enough with matrices to perform the calculations.
Best Answer
The definition of the determinant is a sum of sign weighted products of matrix elements (it seems from the top of the included image). Show that in each of these products exactly one element of row $l$ appears. When multiplying by $\lambda$ this means that every product aquires exactly a single factor $\lambda$. Use distributivity and conclude your argument.