Question

If $A$ is a real skew-symmetric matrix such that $A^2+I=O$, then

• A. $A$ is a square matrix of even order with $\left|A\right|=\pm 1$
• B. $A$ is a square matrix of odd order with $\left|A\right|=\pm 1$
• C. $A$ can be a square matrix of any order with $\left|A\right|=\pm 1$
• D. $A$ is a skew-symmetric matrix of even order with $\left|A\right|=1$

Answer 1

The correct option is D $A$ is a skew-symmetric matrix of even order with $\left|A\right|=1$

Given A is a real skew-symmetric matrix.
$A^{\prime }=-A$
We know that $|A|=|A^{\prime }|$
$|kA|=k^n|A|$
$\therefore |A^{\prime }|=(-1{)}^n|A|$
$\Rightarrow |A|=(-1{)}^n|A|$
$\Rightarrow (1-(-1{)}^n)|A|=O$
$\Rightarrow |A|=0or1-(-1{)}^n=0$ (if $n$ is even)
But given $A^2+I=O$
$\Rightarrow A^2=-I$
$\Rightarrow |A{|}^2=(-1{)}^n|I|=(-1{)}^n\ne 0$
Hence, A is of even order.
$|A|=1$
Hence, option 'D' is correct.