Question

answer question in attachment
help me please .
Urgently needed .
Will mark u brainliest
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Answer 1

$\large \bf \clubs \:  Given :-$

$\sf{2A + 2B = \left[\begin{array}{cc}2&  - 1& 4 \\3 & 2 & 5\end{array}\right] } \\  \\ \sf{A + 2B = \left[\begin{array}{cc}5 & 0 & 3 \\1 & 6 & 2 \end{array}\right]}$

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$\large \bf \clubs \:  To \: Find :-$

• Value of 2B .

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$\large \bf \clubs \:  Solution:-$

We Have,

$\pmb{2A + 2B = \left[\begin{array}{cc}2&  - 1& 4 \\3 & 2 & 5\end{array}\right] } \:  \:  -  -  -  - (1)\\  \\  \:  \:   \:  \:  \:\pmb{A + 2B = \left[\begin{array}{cc}5 & 0 & 3 \\1 & 6 & 2 \end{array}\right]} \:  \: -  -  -  - (2)$

Multiplying Eq. (2) by 2 We Get :

$\pmb{2A + 4B = \left[\begin{array}{cc}10 & 0 & 6 \\2 & 12 & 4\end{array}\right]} \:  \:  -  -  -  - (3)$

Subtracting (1) From (3) We Get :

$\sf \cancel{2A} + 4B -  \cancel{2A}  -2B  \\  = \left[\begin{array}{cc}10 & 0 & 6 \\ 2 & 12 & 4\end{array}\right] - \left[\begin{array}{cc}2 &  - 1 & 4 \\3 & 2 & 5\end{array}\right] \\  \\ :\longmapsto\sf2B = \left[\begin{array}{cc} 10- 2&   0+ 1& 6-  4 \\ 2- 3 &  12- 2 & 4 - 5\end{array}\right]$

$\purple{ \large :\longmapsto  \pmb{ \underline {\boxed{{2B = \left[\begin{array}{cc} 8&   1& 2 \\ - 1 & 10 &  - 1\end{array}\right]} }}}}$

$\underline{\underline{\Large\pink{\mathfrak{  \text{H}ence \:\:option\:\: B\:\: Correct} }}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Given :-

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

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

To Find :-

2B

Solution :-

Multiply A + 2B with 2

$\sf A+2B\times2 = \left[\begin{array}{ccc}5&0&3\\1&6&2\end{array}\right] \times 2$

$\sf 2A+4B=\left[\begin{array}{ccc}10&0&6\\2&12&4\end{array}\right]$

Now, Add both

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

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

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