For the matrix $A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix}$,which of the following is correct?

Let $I$ denote the $3 \times 3$ identity matrix and $P$ be a matrix obtained by rearranging the columns of $I$. Then,

If $A+2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A-B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$,then $\operatorname{tr}(A)-\operatorname{tr}(B) =$

If $A = [2]$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$,then $(BA)' = $ . . . . . . .

If $R(t) = \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}$,then $R(s) \cdot R(t) = $

Let $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$,then: