Question

Assertion :If A is a skew symmetric matrix of odd order, then det $\left(A\right)=0$ Reason: For every square matrix A $det\left(A\right)=det\left(A^{\prime }\right)=det(-A^{\prime }).$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A),
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A)is true but (R} is false,
• D. (A)is false but (R} is true.

Answer 1

The correct option is C (A)is true but (R} is false,

As A is skew symmetric matrix of odd order Let $A=\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 3\\ -2& -3& 0\end{array}\right]$ $\therefore A=-A^{\prime }$ --------(As A is skew symmetric)
$\Rightarrow \left|A\right|=|-A^{\prime }|\Rightarrow \left|A\right|=-\left|A^{\prime }\right|$
$\Rightarrow \left|A\right|=-\left|A\right|\Rightarrow 2\left|A\right|=0$
$\therefore \left|A\right|=0$
But det $\left(A^{\prime }\right)=det(-A^{\prime })$ is not true in general.
$\therefore det(-A^{\prime })=(-1{)}^{3}det(A^{\prime })=-detA$
$\therefore$ Assertion (A) is true & Reason (R) false.
Hence, option C.