Matrices — Class 12 Maths Solution

We have, $A = \left[ {\begin{array}{llllllllllllllllllll}1&2\\4&1\\5&6\end{array}} \right]$

and $B = \left[ {\begin{array}{llllllllllllllllllll}1&2\\6&4\\7&3\end{array}} \right]$
(i) $\therefore$ $(2A + B) = \left[ {\begin{array}{cccccccccccccccccccc}2&4\\8&2\\{10}&{12}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}1&2\\6&4\\7&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&6\\{14}&6\\{17}&{15}\end{array}} \right]$

and ${(2A + B)^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}3&{14}&{17}\\6&6&{15}\end{array}} \right]$

$2{A^\prime } + {B^\prime } = 2\left[ {\begin{array}{llllllllllllllllllll}1&4&5\\2&1&6\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}1&6&7\\2&4&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&{14}&{17}\\6&6&{15}\end{array}} \right] = {(2A + B)^\prime }$

(ii) $(A - B) = \left[ {\begin{array}{llllllllllllllllllll}1&2\\4&1\\5&6\end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll}1&2\\6&4\\7&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0\\{ - 2}&{ - 3}\\{ - 2}&3\end{array}} \right]$

and ${(A - B)^\prime } = \left[ {\begin{array}{llllllllllllllllllll}0&{ - 2}&{ - 2}\\0&{ - 3}&{\,\,\,3}\end{array}} \right]$

${A^\prime } - {B^\prime } = \left[ {\begin{array}{llllllllllllllllllll}1&4&5\\2&1&6\end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll}1&6&7\\2&4&3\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 2}&{ - 2}\\0&{ - 3}&3\end{array}} \right]$
$= {(A - B)^\prime }$