Question:
Using properties of determinants, prove that $\left|\begin{array}{ccc}1 & 1 & 1+3 x \\ 1+3 y & 1 & 1 \\ 1 & 1+3 z & 1\end{array}\right|=9(3 x y z+x y+y z+z x)$.
Solution:
Let $\Delta=\left|\begin{array}{ccc}1 & 1 & 1+3 x \\ 1+3 y & 1 & 1 \\ 1 & 1+3 z & 1\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$
$\Rightarrow \Delta=\left|\begin{array}{ccc}1 & 1 & 1+3 x \\ 3 y & 0 & -3 x \\ 0 & 3 z & -3 x\end{array}\right|$
Expanding along $R_{1}$, we get
$\Delta=1(0+9 x z)-1(-9 x y-0)+(1+3 x)(9 y z-0)$
$=9 x z+9 x y+9 y z+27 x y z$
$=9(3 x y z+x y+y z+z x)$