## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$4$
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. $D=\begin{vmatrix}{x}&{y}&{z}\\{u}&{v}&{w}\\{1}&{2}&{3}\end{vmatrix}=4$ 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, $D_2=-D_1=-(-D)=-(-4)=4$ Subtract $R_3$ from $R_2$ of $D_{B}$ and apply property -3 to obtain: $D_{C}=\begin{vmatrix}{1}&{2}&{3}\\{x-u}&{y-v}&{z-w}\\{u}&{v}&{w}\end{vmatrix}$ So, $D_{C}=D=4$