Answer
$8$
Work Step by Step
In order to work out with the given problem, we will apply some of the following properties of a determinant:
1. The sign of a determinant gets changed when any two columns or two rows are interchanged.
2. When any column or row of a determinant is multiplied by a non-zero number $a$, then the value of the determinant is also multiplied by a non-zero number $a$.
3. When any column or row of a determinant is multiplied by a non-zero number $a$, and then we add it to another column or row, then the value of the determinant does not change.
Thus, we have:
$D=\begin{vmatrix}{x}&{y}&{z}\\{u}&{v}&{w}\\{1}&{2}&{3}\end{vmatrix}=4$
Interchange $R_1$ and $R_3$ to obtain:
$D_{A}=\begin{vmatrix}{1}&{2}&{3}\\{u}&{v}&{w}\\{x}&{y}&{z}\end{vmatrix}=-D=-4$
Interchange $R_2$ and $R_3$ of $D_{A}$ to obtain:
$D_{B}=\begin{vmatrix}{1}&{2}&{3}\\{x}&{y}&{z}\\{u}&{v}&{w}\end{vmatrix}$
So, $D_B=-D_A=-(-D)=-(-4)=4$
Multiply $R_1$ by $-3$ and add it to $R_2$ of $D_{B}$
$D_{C}=\begin{vmatrix}{1}&{2}&{3}\\{x-3}&{y-6}&{z-9}\\{u}&{v}&{w}\end{vmatrix}$
So, $D_{C}=D_B=4$ (we see no change in the determinant)
So, finally multiply $R_3$ by $2$ and add it to $R_2$ of $D_{B}$
$D_{D}=\begin{vmatrix}{1}&{2}&{3}\\{x-3}&{y-6}&{z-9}\\{2u}&{2v}&{2w}\end{vmatrix}$
Thus, we have
$D_{D}=2 D_C=(2)(4)=8$
