Question

If A is a skew-symmetric matrix of order 3,then prove that det A=0.

Answer 1

As A is a skew-symmetric matrix of order 3.So,$A=-A^T$

Now $|A|=|-A^T|=(-1{)}^3|A^T|=-|A|$ $\{\because |kA|=k^n|A|$ where n is order of A;|A|=$|A^T|$.

$\Rightarrow \left|A\right|+\left|A\right|=0$ $\therefore$ $\text{|A|=0 or,det A=0}$