Using the property of determinant and without expanding prove that \(\left| {\begin{array}{*{20}{c}} 0&a&{ - b} \\ { - a}&0&{ - c} \\ b&c&0 \end{array}} \right| = 0\)
Answer
Let \(\Delta = \left| {\begin{array}{*{20}{c}} 0&a&{ - b} \\ { - a}&0&{ - c} \\ b&c&0 \end{array}} \right|\)
[Taking (-1) common from each row]
\(\Rightarrow \Delta = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}} 0&{ - a}&b \\ a&0&c \\ { - b}&{ - c}&0 \end{array}} \right|\)
Interchanging rows and columns in the determinant on R.H.S.,
\(\Delta = - \left| {\begin{array}{*{20}{c}} 0&a&{ - b} \\ { - a}&0&{ - c} \\ b&c&0 \end{array}} \right|\)
\( \Rightarrow \Delta = - \Delta \)
\( \Rightarrow \Delta + \Delta = 0\)
\( \Rightarrow 2\Delta = 0\)
\( \Rightarrow \Delta = 0\) Proved.
[Taking (-1) common from each row]
\(\Rightarrow \Delta = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}} 0&{ - a}&b \\ a&0&c \\ { - b}&{ - c}&0 \end{array}} \right|\)
Interchanging rows and columns in the determinant on R.H.S.,
\(\Delta = - \left| {\begin{array}{*{20}{c}} 0&a&{ - b} \\ { - a}&0&{ - c} \\ b&c&0 \end{array}} \right|\)
\( \Rightarrow \Delta = - \Delta \)
\( \Rightarrow \Delta + \Delta = 0\)
\( \Rightarrow 2\Delta = 0\)
\( \Rightarrow \Delta = 0\) Proved.
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