For $3 \times 3$ matrices $M$ and $N$,which of the following statement$(s)$ is (are) $NOT$ correct?

If the determinant of a $3^{\text{rd}}$ order matrix $A$ is $K$,then the sum of the determinants of the matrices $(AA^T)$ and $(A-A^T)$ is

Let $A = [a_{ij}]$ be a $3 \times 3$ matrix,where
$a_{ij} = 1$,if $i = j$
$a_{ij} = -x$,if $|i - j| = 1$
$a_{ij} = 2x + 1$,otherwise
Let a function $f: R \rightarrow R$ be defined as $f(x) = \det(A)$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to:

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & {\sin \theta } & {\cos \theta } \\ {\sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & {\sin 2\theta } & {\cos 2\theta } \\ {\sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \end{array}} \right|$,$x \ne 0$; then for all $\theta \in \left( {0, \frac{\pi }{2}} \right)$:

Let $A=\left[\begin{array}{ll}2 & 3 \\ a & 0\end{array}\right], a \in R$ be written as $P+Q$ where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If $\operatorname{det}(Q)=9$,then the modulus of the sum of all possible values of the determinant of $P$ is equal to: