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If $a+b+c=0$ then find the value of determinant $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Solution:
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Verified Answer
We have
$$
\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|
$$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$
$$
=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|
$$
Take $(a+b+c)$ common from $R_1$
$$
\begin{aligned}
&=(a+b+c)\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right| \\
&=0 \quad(\because a+b+c=0)
\end{aligned}
$$
$$
\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|
$$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$
$$
=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|
$$
Take $(a+b+c)$ common from $R_1$
$$
\begin{aligned}
&=(a+b+c)\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right| \\
&=0 \quad(\because a+b+c=0)
\end{aligned}
$$
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