Question

Solve this question of matrices in Attachment

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Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:2A + 2B = \begin{gathered}\sf \left[\begin{array}{ccc}2& - 1&4\\3&2&5\end{array}\right]\end{gathered}$

and

$\rm :\longmapsto\:A + 2B = \begin{gathered}\sf \left[\begin{array}{ccc}5& 0&3\\1&6&2\end{array}\right]\end{gathered}$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \bf{ \: 2B}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:2A + 2B = \begin{gathered}\sf \left[\begin{array}{ccc}2& - 1&4\\3&2&5\end{array}\right]\end{gathered} - - - (1)$

and

$\rm :\longmapsto\:A + 2B = \begin{gathered}\sf \left[\begin{array}{ccc}5& 0&3\\1&6&2\end{array}\right]\end{gathered} - - - (2)$

Now, multiply equation (2) by 2, we get

$\rm :\longmapsto\:2A + 4B =2 \begin{gathered}\sf \left[\begin{array}{ccc}5& 0&3\\1&6&2\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:2A + 4B = \begin{gathered}\sf \left[\begin{array}{ccc}10& 0&6\\2&12&4\end{array}\right]\end{gathered} - - - (3)$

Now, Subtracting equation (2) from equation (3), we get

$\rm :\longmapsto\:2B = \begin{gathered}\sf \left[\begin{array}{ccc}10& 0&6\\2&12&4\end{array}\right]\end{gathered} - \begin{gathered}\sf \left[\begin{array}{ccc}2& - 1&4\\3&2&5\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:2B = \begin{gathered}\sf \left[\begin{array}{ccc}10 - 2& 0 + 1&6 - 4\\2 - 3&12 - 2&4 - 5\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:2B = \begin{gathered}\sf \left[\begin{array}{ccc}8&1&2\\ - 1&10& - 1\end{array}\right]\end{gathered}$

Hence,

• Option (B) is correct.

Additional Information :-

1. Matrix addition is possible only when order of matrix are same otherwise its meaningless.

2. Matrix subtraction is possible only when order of matrix are same otherwise its meaningless.

3. Matrix multiplication is defined when number of columns of pre multiplier is equal to the number of rows of post multiplier.