# 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) $\left[ {\begin{array}{*{20}{c}} 8&1&2 \\ 1&{10}&1 \end{array}} \right]$

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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.

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,
$\Rightarrow {\text{2}}\left( {{\text{A + 2B}}} \right){\text{ = 2}}\left[ {\begin{array}{*{20}{c}} 5&0&3 \\ 1&6&2 \end{array}} \right] \\ \Rightarrow 2{\text{A + 4B = }}\left[ {\begin{array}{*{20}{c}} {10}&0&6 \\ 2&{12}&4 \end{array}} \right] \\$
$\Rightarrow 2{\text{A + 4B - }}\left( {2{\text{A + 3B}}} \right) = \left[ {\begin{array}{*{20}{c}} {10}&0&6 \\ 2&{12}&4 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&4 \\ 3&2&5 \end{array}} \right] \\ \Rightarrow {\text{2A + 4B - 2A - 3B = }}\left[ {\begin{array}{*{20}{c}} {10 - 2}&{0 - \left( { - 1} \right)}&{6 - 4} \\ {2 - 3}&{12 - 2}&{4 - 5} \end{array}} \right] \\$
$\Rightarrow {\text{B = }}\left[ {\begin{array}{*{20}{c}} 8&1&2 \\ { - 1}&{10}&{ - 1} \end{array}} \right]$
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.