Using properties of determinants,

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

$\Delta=1(0+9 x z)-1(-9 x y-0)+(1+3 x)(9 y z-0)$

$=9 x z+9 x y+9 y z+27 x y z$

$=9(3 x y z+x y+y z+z x)$

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