Question

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]

Answer 1

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.