1. Given condition

We are given a real $2\times 2$ matrix $A$ such that $A^2=I.$ This means $A$ satisfies the polynomial $x^2-1=(x-1)(x+1)=0.$

So the eigenvalues of $A$ must be among $1$ and $-1$.

2. Analyze Statement-1

Statement-1 says:

If $A\neq I$ and $A\neq -I$, then $\det(A)=-1$.

Let the eigenvalues of $A$ be $\lambda_1,\lambda_2$. Since $A^2=I$, $\lambda_1^2=1,\qquad \lambda_2^2=1,$ so each eigenvalue is either $1$ or $-1$.

Now for a $2\times 2$ matrix, possible eigenvalue pairs are:

• $(1,1)$
• $(-1,-1)$
• $(1,-1)$

If eigenvalues are $(1,1)$, then since the minimal polynomial divides $(x-1)$ or $(x-1)^2$, but $A^2=I$ and over reals the polynomial $x^2-1$ has distinct roots, $A$ is diagonalizable. Hence $A=I$.

Similarly, if eigenvalues are $(-1,-1)$, then $A=-I$.

But given $A\neq I$ and $A\neq -I$, the only remaining possibility is that eigenvalues are $1$ and $-1$.

Therefore, $\det(A)=\lambda_1\lambda_2=(1)(-1)=-1.$

So Statement-1 is true.

3. Analyze Statement-2

Statement-2 says:

If $A\neq I$ and $A\neq -I$, then $\operatorname{tr}(A)\neq 0$.

From the above, in this case the eigenvalues are $1$ and $-1$. Therefore, $\operatorname{tr}(A)=\lambda_1+\lambda_2=1+(-1)=0.$

So actually, $\operatorname{tr}(A)=0,$ not $\neq 0$.

Hence Statement-2 is false.

4. Choose the correct option

• Statement-1: True
• Statement-2: False

Therefore the correct option is $\boxed{\text{D}}.$

5. Comparison with stored answer

Stored correct answer: $\text{D}$

Our derived answer: $\text{D}$

They match.