Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
$\Delta = \left| {\begin{array}{*{20}{r}}
2&{ - 3}&5 \\
6&0&4 \\
1&5&{ - 7}
\end{array}} \right| = - 28{\text{ }}$
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.
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