Practicing Success
If A is a real skew-symmetric matrix such that $A^2+1=O$, then |
A is a square matrix of even order with $|A| = ± 1$ A is a square matrix of odd order with $|A| = ± 1$ A can be a square matrix of any order with $|A| = ± 1$ A is a skew-symmetric matrix of even order with $|A| = ± 1$ |
A is a skew-symmetric matrix of even order with $|A| = ± 1$ |
We have, $A^2+1=O$ $A^2=-I= |A^2|=|I|⇒ |A|^2 =1⇒ |A|=±1$ Let the order of A be $n×n$. Since A is skew-symmetric matrix. $∴A^T=- A$ $⇒|A^T|=|-A|$ $⇒ |A|=(-1)^n |A|$ $⇒|A|=-|A|$, if n is odd $⇒ |A|=0$, if n is odd. But, $|A|≠0$. So, A cannot be a skew-symmetric matrix of odd order. Infact, the determinant of a skew-symmetric matrix of even order is a perfect square. $∴|A|=1$ |