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By using properties of determinants, show that
$\left|\begin{array}{ccc}-a^2 & a b & a c \\ b a & -b^2 & b c \\ a c & c b & -c^2\end{array}\right|=4 a^2 b^2 c^2$
$\left|\begin{array}{ccc}-a^2 & a b & a c \\ b a & -b^2 & b c \\ a c & c b & -c^2\end{array}\right|=4 a^2 b^2 c^2$
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Verified Answer
Taking a, b, c common from $\mathrm{R}_1, \mathrm{R}_2 \& \mathrm{R}_3$ respectively.
$\therefore \quad \Delta=a b c\left|\begin{array}{ccc}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right|$
Again a, b, c are taking common from $\mathrm{C}_1, \mathrm{C}_2 \& \mathrm{C}_3$ Respectively
$\therefore \Delta=\mathrm{a}^2 \mathrm{~b}^2 \mathrm{c}^2\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$ Now Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$
$\begin{aligned}
&\Delta=a^2 b^2 c^2\left|\begin{array}{ccc}
0 & 0 & 2 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right| \\
&=a^2 b^2 c^2 \cdot 2(1+1)=4 a^2 b^2 c^2
\end{aligned}$
$\therefore \quad \Delta=a b c\left|\begin{array}{ccc}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right|$
Again a, b, c are taking common from $\mathrm{C}_1, \mathrm{C}_2 \& \mathrm{C}_3$ Respectively
$\therefore \Delta=\mathrm{a}^2 \mathrm{~b}^2 \mathrm{c}^2\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$ Now Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$
$\begin{aligned}
&\Delta=a^2 b^2 c^2\left|\begin{array}{ccc}
0 & 0 & 2 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right| \\
&=a^2 b^2 c^2 \cdot 2(1+1)=4 a^2 b^2 c^2
\end{aligned}$
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