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?
 A: the original matrix is better. Subtract the identity matrix, the result is the rank one  symmetric $v v^T,$  where $v^T = (x,y,z).$  The vector $v$ is also an eigenvector as $v v^T v  = v (v^T v), $  so that the eigenvalue is $x^2 + y^2 + z^2.$  The other two eigenvalues are $0.$
Now add in the identity matrix, this adds $1$ to each eigenvalue, now the eigenvalues are $1,1,1+x^2 + y^2 + z^2,$  while the determinant is  $1+x^2+y^2+z^2$
A: *

*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.
A: Here is how I think of it, we can think of such  a determinant as a scalar triple product of three vector functions. Explicitly speaking:
$$ D(x,y)= \vec{f}(x,y) \cdot \vec{h}(x,y) \times \vec{g}(x,y)$$
You can imagine three vector fields in space, and the above expression takes those three vectors corresponding to that vector field at a point and evaluates their scalar triple product.
Now, if we have a factorable vector field meaning something of the form:
$$ \vec{f}(x,y) = h(x,y) \vec{q}(x,y)$$
That is a vector field which can be written in terms of a 'modulating scale factor' which varies from point to point times another a vector field, then, we can pull of these scale factors from the scalar triple product and just take the scalar product of the 'factored' vector functions.
Hopefully that makes sense.
