question_answer

Show that the determinant value of a skew-symmetric matrix of odd order is always zero.

Answer:

Let A be a skew-symmetric matrix of order n. Then,    \[A'=-A\] \[\Rightarrow \] \[|A'|\,\,=\,\,|-\,A|\] \[\Rightarrow \]   \[|A'|\,\,={{(-1)}^{n}}|A|\] \[\Rightarrow \]   \[|A'|\,\,=-|A|\]            [\[\because \] n is odd] \[\Rightarrow \]   \[|A|\,\,=-|A|\]         \[[\because \,\,\,|A'|\,\,=-|A|]\] \[\Rightarrow \]   \[2|A|\,\,=0\] \[\Rightarrow \]   \[|A|\,\,=0\] Hence, determinant value of a skew-symmetric matrix of odd order is always zero.