$\Delta  = \left| {\begin{array}{*{20}{r}}
  2&{ – 3}&5 \\ 
  6&0&4 \\ 
  1&5&{ – 7} 
\end{array}} \right| =  – 28{\text{ }}$

Interchanging rows $\mathrm{R}_{2}$ and $\mathrm{R}_{3}$ i.e., $\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3},$ we have

${\Delta _1} = \left| {\begin{array}{*{20}{r}}
  2&{ – 3}&5 \\ 
  1&5&{ – 7} \\ 
  6&0&4 
\end{array}} \right|$

Expanding the determinant $\Delta_{1}$ along first row, we have

${\Delta _1} = 2\left| {\begin{array}{*{20}{c}}
  5&{ – 7} \\ 
  0&4 
\end{array}} \right| – ( – 3)\left| {\begin{array}{*{20}{c}}
  1&{ – 7} \\ 
  6&4 
\end{array}} \right| + 5\left| {\begin{array}{*{20}{c}}
  1&5 \\ 
  6&0 
\end{array}} \right|$

$ = 2(20 – 0) + 3(4 + 42) + 5(0 – 30)$

$ = 40 + 138 – 150 = 28$

Clearly $\quad \Delta_{1}=-\Delta$

Hence, Property $2$ is verified.