Question

Matrices - Find A and B

Answer 1

Answer:

sub B in given equation

we get A=bracket 3 -1

1 19by5

Answer 2

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

Given that

$\rm :\longmapsto\:2A + B = \bigg[ \begin{matrix}3& - 4 \\ 2&9 \end{matrix} \bigg] - - - (1)$

and

$\rm :\longmapsto\:A - 2B = \bigg[ \begin{matrix}9&3 \\ 1&1 \end{matrix} \bigg] - - - (2)$

On multiply equation (1) by 2, we get

$\rm :\longmapsto\:4A + 2B = \bigg[ \begin{matrix}6& - 8 \\ 4&18 \end{matrix} \bigg] - - - (3)$

Now, on adding equation (2) and (3), we get

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}6& - 8 \\ 4&18 \end{matrix} \bigg] + \bigg[ \begin{matrix}9&3 \\ 1&1 \end{matrix} \bigg]$

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}6 + 4& - 8 + 18 \\ 4 + 1&18 + 1 \end{matrix} \bigg]$

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}10& 10 \\ 5&19 \end{matrix} \bigg]$

$\bf :\longmapsto\:A= \bigg[ \begin{matrix}2& 2 \\ 1& \dfrac{19}{5} \end{matrix} \bigg]$

On substituting the value of A, in equation (1), we get

$\rm :\longmapsto\:\bigg[ \begin{matrix}4& 4 \\ 2& \dfrac{38}{5} \end{matrix} \bigg]+ B = \bigg[ \begin{matrix}3& - 4 \\ 2&9 \end{matrix} \bigg]$

$\rm :\longmapsto\:B = \bigg[ \begin{matrix}3& - 4 \\ 2&9 \end{matrix} \bigg] - \bigg[ \begin{matrix}4& 4 \\ 2& \dfrac{38}{5} \end{matrix} \bigg]$

$\rm :\longmapsto\:B = \bigg[ \begin{matrix} - 1& - 8 \\ 0& \dfrac{7}{5} \end{matrix} \bigg]$

Hence,

$\purple{\bf :\longmapsto\:A= \bigg[ \begin{matrix}2& 2 \\ 1& \dfrac{19}{5} \end{matrix} \bigg] }$

and

$\purple{\rm :\longmapsto\:B = \bigg[ \begin{matrix} - 1& - 8 \\ 0& \dfrac{7}{5} \end{matrix} \bigg]}$

Additional Information :-

1. Matrix addition of two matrices A and B is possible only when order of both the matrices A and B are same.

2. Matrix subtraction of two matrices A and B is possible only when order of both the matrices A and B are same.

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