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By using properties of determinants, show that
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$
Solution:
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Verified Answer
Let $\Delta=\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \quad \ldots (i)$
Multiplying each column by $-1$
$\Delta=(-1)^3\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
b & c & 0
\end{array}\right|=-\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \quad \ldots(ii)$
Also rearranging rows and columns
$\Delta=\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \quad \ldots(iii)$
Adding (ii) and (iii)
$\begin{aligned}
&2 \Delta=-\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right|+\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \\
&\Rightarrow \quad 2 \Delta=0 \Rightarrow \Delta=0
\end{aligned}$
Multiplying each column by $-1$
$\Delta=(-1)^3\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
b & c & 0
\end{array}\right|=-\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \quad \ldots(ii)$
Also rearranging rows and columns
$\Delta=\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \quad \ldots(iii)$
Adding (ii) and (iii)
$\begin{aligned}
&2 \Delta=-\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right|+\left|\begin{array}{ccc}
0 & -a & b \\
a & 0 & c \\
-b & -c & 0
\end{array}\right| \\
&\Rightarrow \quad 2 \Delta=0 \Rightarrow \Delta=0
\end{aligned}$
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