Question

Topic Matrices

solve the question in attachment​

Answer 1

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

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

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

• 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 2

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 :

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

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

Correct Option = B