Scalar "exchange" in determinants Consider the following:

Find
$$
    \begin{vmatrix}
    x^2+1 & xy & xz \\
    xy & y^2+1 & yz \\
    xz & yz & z^2+1 \\
    \end{vmatrix}
$$
Multiplying $R_1,R_2,R_3$ with $x,y,z$ and dividing $C_1,C_2,C_3$ by $x,y,z$ respectively, we get
$$
    \begin{vmatrix}
    x^2+1 & x^2 & x^2 \\
    y^2 & y^2+1 & z^2 \\
    z^2 & z^2 & z^2+1 \\
    \end{vmatrix}
$$

I have seen this technique of distributing scalars and taking them out being used extensively to simplify determinants quickly.
Is this a known "trick" in linear algebra? What is the intuition behind this? Are there some specific patterns of determinants where this technique proves helpful?

*

*Multiplying the $k$th row by $c$ is accomplished by multiplying on the left by the matrix $M(k,c)$ given by
$$ M(k,c)_{ij}=\begin{cases} c, & i=j=k,\\ 1, &i=j\neq k, \\ 0 &\text{otherwise} \end{cases}$$


*Multiplying the $k$th column by $c$ is done by multiplying by $M(k,c)$ on the right.


*Determinants of products are products of determinants: $|AB|=|A| |B|$, more generally $|A_1\dots A_n| = |A_1|\dots |A_n|$.


*Let $$ A=\begin{bmatrix}
    x^2+1 & xy & xz \\
    xy & y^2+1 & yz \\
    xz & yz & z^2+1 \\
    \end{bmatrix} , B=\begin{bmatrix}
    x^2+1 & x^2 & x^2 \\
    y^2 & y^2+1 & z^2 \\
    z^2 & z^2 & z^2+1 \\
    \end{bmatrix}$$
Then points 1 and 2 let us translate the given algorithm into $$B=M(1,x)M(2,y)M(3,z)AM(1,\frac1x)M(2,\frac1y)M(3,\frac1z)$$
since $|M(1,x)| |M(1,1/x)|=1$, the result follows from point 3.
