If \[2{\text{A + 3B = }}\left[ {\begin{array}{*{20}{c}}
  2&{ - 1}&4 \\
  3&2&5
\end{array}} \right]\] and \[{\text{A + 2B = }}\left[ {\begin{array}{*{20}{c}}
  5&0&3 \\
  1&6&2
\end{array}} \right]\] then find B-
A)$\left[ {\begin{array}{*{20}{c}}
  8&{ - 1}&2 \\
  { - 1}&{10}&{ - 1}
\end{array}} \right]$ B)$\left[ {\begin{array}{*{20}{c}}
  8&1&2 \\
  { - 1}&{10}&{ - 1}
\end{array}} \right]$ C) $\left[ {\begin{array}{*{20}{c}}
  8&1&{ - 2} \\
  { - 1}&{10}&{ - 1}
\end{array}} \right]$ D) $\left[ {\begin{array}{*{20}{c}}
  8&1&2 \\
  1&{10}&1
\end{array}} \right]$

Hint: To find the value of B, first multiply \[{\text{A + 2B = }}\left[ {\begin{array}{*{20}{c}}
  5&0&3 \\
  1&6&2
\end{array}} \right]\] by 2 on both sides then subtract the given \[2{\text{A + 3B = }}\left[ {\begin{array}{*{20}{c}}
  2&{ - 1}&4 \\
  3&2&5
\end{array}} \right]\] from the obtained multiplication result. You will get the value of B.

Complete step by step answer:

 Given, \[2{\text{A + 3B = }}\left[ {\begin{array}{*{20}{c}}
  2&{ - 1}&4 \\
  3&2&5
\end{array}} \right]\] ---- (i)
And \[{\text{A + 2B = }}\left[ {\begin{array}{*{20}{c}}
  5&0&3 \\
  1&6&2
\end{array}} \right]\]----- (ii)
Here, the matrices given are of order $\left( {2 \times 3} \right)$ .We have to find B. First we multiply eq. (ii) by 2 on both sides,
\[
   \Rightarrow {\text{2}}\left( {{\text{A + 2B}}} \right){\text{ = 2}}\left[ {\begin{array}{*{20}{c}}
  5&0&3 \\
  1&6&2
\end{array}} \right] \\
   \Rightarrow 2{\text{A + 4B = }}\left[ {\begin{array}{*{20}{c}}
  {10}&0&6 \\
  2&{12}&4
\end{array}} \right] \\
 \]

Now on subtracting eq. (i) from the obtained result, we have
$
   \Rightarrow 2{\text{A + 4B - }}\left( {2{\text{A + 3B}}} \right) = \left[ {\begin{array}{*{20}{c}}
  {10}&0&6 \\
  2&{12}&4
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  2&{ - 1}&4 \\
  3&2&5
\end{array}} \right] \\
   \Rightarrow {\text{2A + 4B - 2A - 3B = }}\left[ {\begin{array}{*{20}{c}}
  {10 - 2}&{0 - \left( { - 1} \right)}&{6 - 4} \\
  {2 - 3}&{12 - 2}&{4 - 5}
\end{array}} \right] \\
 $
On solving the above expression, we get
$ \Rightarrow {\text{B = }}\left[ {\begin{array}{*{20}{c}}
  8&1&2 \\
  { - 1}&{10}&{ - 1}
\end{array}} \right]$
Hence, the correct answer is ‘B’.

Note: Here, the order of matrices is $\left( {2 \times 3} \right)$ because order=number of rows × number of columns. Also, since the order of both matrices [the obtained one and eq.(i)] is the same, so we can subtract the elements in the same position easily and obtain the matrix of the same order.