Assertion :Let $A$ be a $2\times 2$ matrix with real entries. Let $I$ be the $2\times 2$ identity matrix. Denote by $tr\left(A\right)$, the sum of diagonal entries of $A$. Assume that $A^2=I$.

If $A\ne I$ and $A\ne -I$, then $det\left(A\right)=-1$.

Reason: If $A\ne I$ and $A\ne -I$, then $tr\left(A\right)\ne 0$.

The correct option is C Assertion is correct but Reason is incorrect
$LetA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$.
Now, $A^2=I\Rightarrow det\left(A^2\right)=1$
$\Rightarrow (detA{)}^2=1\Rightarrow detA=\pm 1$.
$Also,A^2=I\Rightarrow A=A^{-1}$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\frac{1}{detA}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

If $detA=1$, then
$a=d,b=-b,c=-c\Rightarrow a=d,b=c=0$.
In this case $A=\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]$.
$|A|=1\Rightarrow a^2=1\Rightarrow a=\pm 1$
$\therefore A=I$ or $A=-I$.
A contradiction.
Thus, $det\left(A\right)=-1$
$\therefore \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=-\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]=\left[\begin{array}{cc}-d& b\\ c& -a\end{array}\right]$
$\Rightarrow a=-d\Rightarrow tr\left(A\right)=a+d=0$.
$\therefore$ Statement-1: is true and statement-2 is false.
Hence, option C.