Topic Matricessolve the question in attachment

1.	Topic Matricessolve the question in attachment
Answer» $\large \bf \clubs \: Given :-$ $\sf{2A + 2B = \left[\begin{array}{cc}2& - <klux>1</klux>& 4 \\<klux>3</klux> & 2 & 5\end{array}\right] } \\ \\ \sf{A + 2B = \left[\begin{array}{cc}5 & 0 & 3 \\1 & 6 & 2 \end{array}\right]}$ $\large \bf \clubs \: To \: Find :-$ VALUE of 2B . $\large \bf \clubs \: Solution :-$ 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 : $\sf \cancel{2A} + 4B - \cancel{2A} -2B \\ = \left[\begin{array}{cc}10 & 0 & 6 \\ 2 & 12 & 4\end{array}\right] - \left[\begin{array}{cc}2 & - 1 & 4 \\3 & 2 & 5\end{array}\right] \\ \\ :\longmapsto\sf2B = \left[\begin{array}{cc} 10- 2& 0+ 1& 6- 4 \\ 2- 3 & 12- 2 & 4 - 5\end{array}\right] \\ \\$ $\purple{ \large :\longmapsto \pmb{ \underline {\boxed{{2B = \left[\begin{array}{cc} 8& 1& 2 \\ - 1 & 10 & - 1\end{array}\right]} }}}}$ $\underline{\underline{\Large\pink{\mathfrak{ \text{H}ence \:\:option\:\: B\:\: Correct} }}}$ $\LARGE\red{\mathfrak{ \text{W}hich \:\:is\:\: the\:\: required} }\\ \Huge \red{\mathfrak{ \text{ A}nswer.}}$

Answer»

$\large \bf \clubs \: Given :-$

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

$\large \bf \clubs \: To \: Find :-$

VALUE of 2B .

$\large \bf \clubs \: Solution :-$

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 :

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