Answer
8
Work Step by Step
We know:
i) Switching any two rows or columns causes the determinant to switch sign.
$\implies$ doing this twice causes the determinant to stay the same.
ii) Adding a multiple of any row to another row does not change the determinant value.
iii) Taking the common factor from a row outside the determinant causes the determinant to be multiplied by that common factor.
$\begin{vmatrix}
1&2&3\\
x-3&y-6&z-9\\
2u&2v&2w
\end{vmatrix}$
Performing $R_{3} \leftrightarrow R_{1},$ then $R_{2} \leftrightarrow R_{1}$, by rule (i), the determinant stays the same.
$\implies \begin{vmatrix}
1&2&3\\
x-3&y-6&z-9\\
2u&2v&2w
\end{vmatrix} \rightarrow \begin{vmatrix}
x-3&y-6&z-9\\
2u&2v&2w\\
1&2&3
\end{vmatrix}$
Performing $R_{1}+3\times R_{3}$, by rule (ii), the determinant stays the same.
$\implies \begin{vmatrix}
x-3&y-6&z-9\\
2u&2v&2w\\
1&2&3
\end{vmatrix} \rightarrow
\begin{vmatrix}
x&y&z\\
2u&2v&2w\\
1&2&3
\end{vmatrix}$
Taking out common factor 2 from $R_{2}$, by rule (iii), the determinant is multiplied by 2.
$\implies 2\begin{vmatrix}
x&y&z\\
u&v&w\\
1&2&3
\end{vmatrix} = 2\times 4 = 8$
