prove that

Question:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

Solution:

Given, $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

[Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ]

$=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

[Taking $(a+b+c)$ common from the first row]

$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

[Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $C_{2} \rightarrow C_{2}-C_{3}$ ]

$=(a+b+c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & -(a+b+c) & 2 b \\ a+b+c & a+b+c & c-a-b\end{array}\right|$

Lastly, expanding along R1, we have

= (a + b + c) [1 x 0 + (a + b + c)2]

= (a + b + c)3

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