question_answer

If A is a skew-symmetric matrix of odd order n, then show that |A| = 0.

Answer:

We have, A is a skew-symmetric matrix. \[\therefore \]      \[{{A}^{T}}=-A\] \[\Rightarrow \]   \[|{{A}^{T}}|\,=\,|-A|\] \[\Rightarrow \]   \[|{{A}^{T}}|\,={{(-\,1)}^{n}}|A|\]       \[[\because \,\,\,|KA|\,={{K}^{n}}|A|]\] \[\Rightarrow \]   \[|A|\,={{(-\,1)}^{n}}|A|\]                  \[[\because \,\,\,|{{A}^{T}}|\,=\,|A|]\] \[\Rightarrow \]   \[|A|\,=-|A|\]               \[[\because \,\,\,n\,\,\text{is}\,\,\text{odd}]\] \[\Rightarrow \]   \[2|A|=0\] \[\Rightarrow \] \[|A|\,\,=0\] Hence, the determinant of a skew-symmetric matrix of odd order is zero.