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નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
Solution
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c)\end{array}\right|$
$=(a+b+c)^{3}\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1\end{array}\right|$
Expanding along $C_{3},$ we have:
$\Delta=(a+b+c)^{3}(-1)(-1)=(a+b+c)^{3}$
Hence, the given result is proved.
