Question

Assertion : The determinant of a skew symmetric matrix of even order is perfect square.
Reason : The determinant of a skew symmetric matrix of odd order is equal to zero.

• A. Assertion and reason both are correct and reason is correct explanation of the assertion.
• B. Assertion and reason both are correct but reason is not correct explanation of the assertion.
• C. Assertion is wrong.
• D. Reason is wrong.

Answer 1

Answer: (C)

Assertion is wrong.

Both assertion and reason are correct but reason is not correct explanation of assertion.

Let take examples of both.

Skew-symmetric determinant odd order vanishes.

$\Rightarrow A=\left[\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right]$

$\Rightarrow \left|A\right|=\left|\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right|$

Changing rows into columns, takes $(-1)$ common from three columns.

$\Rightarrow \left|A\right|=(-1{)}^3\left|\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right|$

$\Rightarrow \left|A\right|=(-1{)}^3\left|A\right|$

$\Rightarrow 2\left|A\right|=0$

$\Rightarrow \left|A\right|=0$

Skew symmetric determinant of even under is a perfect square.

$\Rightarrow A=\left[\begin{array}{ccc}0& x& \begin{array}{cc}y& z\end{array}\\ -x& 0& \begin{array}{cc}c& b\end{array}\\ \begin{array}{c}-y\\ -z\end{array}& \begin{array}{c}-c\\ -b\end{array}& \begin{array}{c}\begin{array}{cc}0& a\end{array}\\ \begin{array}{cc}-a& 0\end{array}\end{array}\end{array}\right]$

$\Rightarrow \left|A\right|=\left|\begin{array}{ccc}0& x& \begin{array}{cc}y& z\end{array}\\ -x& 0& \begin{array}{cc}c& b\end{array}\\ \begin{array}{c}-y\\ -z\end{array}& \begin{array}{c}-c\\ -b\end{array}& \begin{array}{c}\begin{array}{cc}0& a\end{array}\\ \begin{array}{cc}-a& 0\end{array}\end{array}\end{array}\right|$

Multiplying column $2$ by 'a' , we get

$\Rightarrow \left|A\right|=\frac{1}{a}\left|\begin{array}{ccc}0& ax& \begin{array}{cc}y& z\end{array}\\ -x& 0& \begin{array}{cc}c& b\end{array}\\ \begin{array}{c}-y\\ -z\end{array}& \begin{array}{c}-ac\\ -ab\end{array}& \begin{array}{c}\begin{array}{cc}0& a\end{array}\\ \begin{array}{cc}-a& 0\end{array}\end{array}\end{array}\right|$

Expanding by column 2, we get

$=\frac{-(ax-by+cz)}{a}\left|\begin{array}{ccc}-x& c& b\\ -y& 0& a\\ -z& -a& 0\end{array}\right|$

$=\frac{(ax-by+cz)}{a}\left|\begin{array}{ccc}x& c& b\\ y& 0& a\\ z& -a& 0\end{array}\right|$

$=\frac{(ax-by+cz)}{a\times a}\left|\begin{array}{ccc}ax& ac& ab\\ y& 0& a\\ z& -a& 0\end{array}\right|$

Hence, the answer is assertion is wrong.