If $A$ and $B$ be two non-singular matrices of order $3$ such that $2A+3BB^{T}=I$ and $B^{-1}=A^{T}$, then determine the value of the following:
(i) $\det\left(B^{T}-2B+3B^{3}+3BA\right)$
(ii) $\det\left(A^{-1}-3B^{3}\right)$
(iii) $\text{trace}\left(A^{-1}+I-AB-3B^{3}\right)$
My Attempt:
Given $2A+3BB^{T}=I$
Taking transpose $2A^{T}+3BB^{T}=I$
So on subtracting we get $A=A^{T}$ i.e. $A$ is symmetric
Given that, $B^{-1}=A^{T}=A$ and thus $AB=BA=I$
Now, all I need to get the answers to the above questions is to somehow prove that $B$ is also symmetric i.e. $B^{T}=B$
$AB=I$
$B^TA^T=I$
$B^TB^{-1}=I$
$B^T=B$
Given that $2A+3BB^T=I$
$2B^{-1}+3B^2=I$
$B-3B^3=2I$
Thus,$B^T-2B+3B^3+3BA=B-2B+3B^3+3I=3B^3-B+3I=-2I+3I=I$
Also, $A^{-1}+I-AB-3B^3=A^{-1}-3B^3=B-3B^3=2I$
I will assume $A,B$ real matrices.
Recall the Spectral Real Theorem:
The matrix $BB^T$ is symmetric so there exists $M\in O(3)$ such that $MBB^TM^T = D$ with $D$ diagonal matrix. Now: \begin{gather} 2A+3BB^T=I\\ M(2A+3BB^T)M^T=I\\ MAM^T = \frac{1}{2}(I-3D) \end{gather} So $A$ is orhogonally diagonalizable and thanks to the spectral theorem it is symmetric.
Edit: This part does not require the real spectral theorem. Take simply the transpose of the first equality in the hypotesis. This show that $A$ is symmetric.
Now we have $B^{-1}=A^T = A$ and hence $B=A^{-1}$ and $B$ is also symmetric. The first hypotesis become: \begin{equation} 2B^{-1}+3B^2=I \end{equation} multiplicand LHS and RHS by $B$ we have: \begin{equation} 3B^3-B+2I=0 \end{equation} Denoting with $m_B(T)$ the minimal polynomial of $B$, we have that $m_B(t)$ divides $(3t^3-t+2)=(t+1)(3t^2-3t+2)$ and, since $B$ is diagonalizable, we have $m_B(t)=(t+1)$. So $B$ has to be the matrix $-I$ and now: \begin{gather} \det(B^T-2B+3B^3+3BA)=\det(-I + 2I - 3I + 3I) = \det(I) = 1\\ \det(A^{-1}-3B^3)=\det(-I+3I) = 2^3\det(I) = 8\\ \text{trace}(A^{-1}+I-AB-3B^3) = \text{trace}(-I+I-I+3I) = \text{trace}(2I) = 6 \end{gather}