Let $n$ be an odd positive integer.
Determine whether there exists an $n \times n$ real matrix $A$ such that
\[A^2+I=O,\]
where $I$ is the $n \times n$ identity matrix and $O$ is the $n \times n$ zero matrix.

If such a matrix $A$ exists, find an example. If not, prove that there is no such $A$.

The key technique to solve this problem very easily is determinant.
Recall the following properties of the determinant.
Let $A, B$ be $n\times n$ matrices and $c$ be a scalar Then we have

$\det(AB)=\det(A)\det(B)$

$\det(cA)=c^n\det(A)$.

Solution.

When $n$ is odd.

When $n$ is odd, we prove that there is no $n \times n$ real matrix $A$ such that $A^2+I=O$.
Seeking a contradiction, assume that we have $A$ such that $A^2+I=O$.
Since we have $A^2=-I$, we have
\[\det(A^2)=\det(-I).\]

Using the properties of determinant, we obtain
\[\det(A)^2=(-1)^n\det(I)=-1\]
because $n$ is odd and $\det(I)=1$.

Since $A$ is a real matrix, the determinant of $A$ is also real.
Thus, $\det(A)^2=-1$ is impossible. Hence there is no such $A$.

When $n$ is even.

On the other hand, if $n$ is even there is $A$ such that $A^2+I=O$.
For example, consider
\[A=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\]
Then a direct computation shows that $A^2=-I$, hence $A^2+I=O$.

Comment.

Recall that the imaginary number $i$ is the number whose square is $-1$.
Similarly, we found above that the square of the matrix $A=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ is $-I$.

But when $n$ is odd, there is no such matrix $A$ as we showed.

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