Question

If $A$ is a real skew-symmetric matrix such that $A^2+I=0$, then which one of the following is true?

• A. $A$ satisfies $A^2=A$ and $A$ is of even order
• B. $A$ satisfies $A^2=I$ and $A$ is of odd order
• C. $A^2=2A$
• D. $A$ satisfies $A{A}^{\prime }=I$ and $A$ is of even order

Answer 1

The correct option is D $A$ satisfies $A{A}^{\prime }=I$ and $A$ is of even order

$A^{\prime }=–A$
$\Rightarrow A{A}^{\prime }=–A^2=I\Rightarrow A{A}^{\prime }=I$
Considering determinant, $|A{A}^{\prime }|=|I|$
$\Rightarrow |A{|}^2=1$
$\Rightarrow |A|=1,–1$
Let $n$ be the order of $A.$
$\Rightarrow |A^{\prime }|=|–A|=(–1{)}^n|A|$
$\Rightarrow |A|1–(–1{)}^n=0\text{and}|A|\ne 0$
$\Rightarrow 1–(–1{)}^n=0\Rightarrow n$ is even.