Let $2A+B = \left[ {\begin{array}{*{20}{c}}
  1&0&3 \\ 
  { – 1}&4&6 \\ 
  2&5&2 
\end{array}} \right],\,A – 2B = \left[ {\begin{array}{*{20}{c}}
  2&{ – 1}&5 \\ 
  0&3&6 \\ 
  1&2&1 
\end{array}} \right]$ . Then $Tr(A) -Tr(B)$ has the value equal to (where $Tr(A)$ denotes the trace of matrix $A$)

If the matrix $\left[ {\begin{array}{*{20}{c}}0&1&{ – 2}\\{ – 1}&0&3\\\lambda &{ – 3}&0\end{array}} \right]$ is singular, then $\lambda $=

Let $A = \left[ {\begin{array}{*{20}{c}}
p&{13}\\
{ – 13}&p
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{4q}&{85}\\
{ – 2}&1
\end{array}} \right]$  where  $p,q \in N$. It is given that $\left| A \right| = \left| B \right|$ and  $p,q \in[1,1000]$. Then total number of ordered pairs $(p,q)$ is

If $A = \left( {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}{ – 5}&7&1\\1&{ – 5}&7\\7&1&{ – 5}\end{array}} \right)$ then $AB$ is equal to

Let $A$ and $B$ be two symmetric matrices of order $3.$

Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices.

Statement $-2:$ $AB$ is symmetric matrix if matrix multiplication of $A$ with $B$ is commutative.