If $AA^T=A^TA=I$ and $\det(A)=1$ then $p_A(1)=0$ If $AA^T=A^TA=I$ and $\det(A)=1$ then $p_A(1)=0$.
Where $A\in M_3(\mathbb{R})$
My approach:
We known from the first equation that $A^T=A^{-1}$ and $\lambda_1\lambda_2\lambda_3=1$. Now,since $A$ and $A^T$ have the same characteristic polynomial, they have the same eigenvalues. We also know that the eigenvalues for $A^{-1}$ are $\frac{1}{\lambda_1},\frac{1}{\lambda_2},\frac{1}{\lambda_3}$ and since $A^T=A^{-1}$ this basically means that $\frac{1}{\lambda_1}=\lambda_1,\frac{1}{\lambda_2}=\lambda_2,\frac{1}{\lambda_3}=\lambda_3$
From here there are 2 possible solutions:
$\lambda_1=\lambda_2=-1,\lambda_3=1$ or $\lambda_1=\lambda_2=\lambda_3=1$
Is my approach correct?
 A: Asserting that $A^TA=AA^T=\operatorname{Id}$ is the same thing as saying that $A$ is an orthogonal matrix. So, for each $v\in\mathbb{R}^3$, $\|Av\|=\|v\|$. In particular, if $v$ is an eigenvector with eigenvalue $\lambda$ (such a $v$ must exist, since $\mathbb{R}^3$ is odd-dimensional), $|\lambda|=1$, that is $\lambda=\pm1$. If $1$ is an eigenvalue,, then $P_A(1)=0$. Suppose otherwise. Then $-1$ is an eigenvalue. Let $W=(\mathbb{R}u)^\perp$. Then $A.W\subset W$ and then $A$ is an orthogonal transformation of $W$ whose determinant is $-1$. But then there is a vector $w\in W\setminus\{0\}$ such that $A.w=w$, that is, $1$ is an eigenvalue of $A$.
A: $A$ is an orthogonal matrix. Hence all the eigenvalues have unit modulus.
Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues of the matrix.
Since, $Det(A)=1\implies\lambda_1\lambda_2\lambda_3=1$------------------(*)
Case-1:
If all $\lambda_i$'s are real, they can take values $\pm1$. 
From (*) it follows that $1$ must be an eigenvalue.
Case-2: 
If not all $\lambda_i$'s are real then we have a pair of complex conjugates.
From (*) it again follows that $1$ must be an eigenvalue.
Thus, $p_A(1)=0$ 
