Question

A square matrix $A$ is said to be orthogonal if $A^{\prime }A=A{A}^{\prime }=I_n$

If $A$ is real skew-symmetric matrix is such that $A^2+I=0$, then

• A. $A$ is orthogonal matrix
• B. $A$ is orthogonal matrix of odd order
• C. $A$ is orthogonal matrix of even order
• D. none of these

Answer 1

Answer: (C)

$A$ is orthogonal matrix of even order

Given A is skew-symmetric matrix and $A^2+I=O$
i.e $A^{\prime }=-A$
$I=-A^2=(-A)A=A^{\prime }A$
$\Rightarrow A$ is orthogonal.
Now, $A^{\prime }A=I_n\Rightarrow det\left(A^{\prime }A\right)=1$
$\Rightarrow det\left(A^{\prime }\right)detA=1\Rightarrow (-1{)}^{n}det(A{)}^2=1$ ($\because det(-A)=(-1{)}^{n}A$)
$\therefore$ A is an orthogonal matrix of even order.
Hence, option C.