Let $A + 2B = \left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  6&{ - 3}&3 \\
  { - 5}&3&1
\end{array}} \right)$ and $2A - B = \left( {\begin{array}{*{20}{c}}
  2&{ - 1}&5 \\
  2&{ - 1}&6 \\
  0&1&2
\end{array}} \right)$. If $Tr\left( A \right)$ denotes the sum of all diagonal elements of the matrix $A$, then $Tr\left( A \right) - Tr\left( B \right)$ has value equal to:
A.$0$
B. $1$
C. $3$
D. $2$

Hint: Write the trace of given matrices using the sum of all diagonal elements then make two equations and find the values of $Tr\left( A \right)$ and $Tr\left( B \right)$ using both the equations. Put the required values in $Tr\left( A \right) - Tr\left( B \right)$.

Formula Used:
Trace formula –
$Tr\left( {mX} \right) + Tr\left( {nY} \right) = mTr\left( X \right) + nTr\left( Y \right)$
Where, $m$ and $n$ are constants.

Complete step by step solution:
 Given that,
$A + 2B = \left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  6&{ - 3}&3 \\
  { - 5}&3&1
\end{array}} \right)$ and $2A - B = \left( {\begin{array}{*{20}{c}}
  2&{ - 1}&5 \\
  2&{ - 1}&6 \\
  0&1&2
\end{array}} \right)$
Here, $Tr\left( {A + 2B} \right) = 1 + \left( { - 3} \right) + 1$
$Tr(A) + 2Tr(B) = - 1$
$Tr(A) = - 1 - 2Tr(B) - - - - - - (1)$
And $Tr\left( {2A - B} \right) = 2 + \left( { - 1} \right) + 2$
$2Tr\left( A \right) - Tr(B) = 3$
From equation (1)
$2\left( { - 1 - 2Tr(B)} \right) - Tr(B) = 3$
$ - 2 - 4Tr(B) - Tr(B) = 3$
$Tr(B) = - 1$
$ \Rightarrow Tr(A) = 1$
$\therefore Tr\left( A \right) - Tr\left( B \right) = 1 - ( - 1) = 2$

Option ‘D’ is correct

Note: The key concept involved in solving this problem is the good knowledge of different methods of solving the system of linear equations. Students must remember how to apply any of the methods (Elimination, Substitution, Graphical) to find the value of unknown variables.