Question

How to find matrice A and B

Answer 1

Step-by-step explanation:

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Answer 2

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

Given matrices are

$\rm :\longmapsto\:2A + B = \bigg[ \begin{matrix}3& - 4 \\ 2&7 \end{matrix} \bigg] - - - (1)$

and

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

Now,

Multiply equation (1) by 2, we get

$\rm :\longmapsto\:4A + 2B = \bigg[ \begin{matrix}6& - 8 \\ 4&14 \end{matrix} \bigg] - - - (3)$

Now, Adding equation (2) and (3), we get

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}6& - 8 \\ 4&14 \end{matrix} \bigg] + \bigg[ \begin{matrix}4&3 \\ 1&1 \end{matrix} \bigg]$

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}6 + 4& - 8 + 3 \\ 4 + 1&14 + 1 \end{matrix} \bigg]$

$\rm :\longmapsto\:5A= \bigg[ \begin{matrix}10& - 5 \\ 5&15 \end{matrix} \bigg]$

$\bf :\longmapsto\:A= \bigg[ \begin{matrix}2& -1 \\ 1&3 \end{matrix} \bigg]$

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

$\rm :\longmapsto\:2\bigg[ \begin{matrix}2& -1 \\ 1&3 \end{matrix} \bigg] + B = \bigg[ \begin{matrix}3& - 4 \\ 2&7 \end{matrix} \bigg]$

$\rm :\longmapsto\:\bigg[ \begin{matrix}4& -2 \\ 2&6 \end{matrix} \bigg] + B = \bigg[ \begin{matrix}3& - 4 \\ 2&7 \end{matrix} \bigg]$

$\rm :\longmapsto\: B = \bigg[ \begin{matrix}3& - 4 \\ 2&7 \end{matrix} \bigg] - \bigg[ \begin{matrix}4& -2 \\ 2&6 \end{matrix} \bigg]$

$\rm :\longmapsto\: B = \bigg[ \begin{matrix}3 - 4& - 4 + 2 \\ 2 - 2&7 - 6 \end{matrix} \bigg]$

$\bf :\longmapsto\: B = \bigg[ \begin{matrix} - 1& -2 \\ 0&1 \end{matrix} \bigg]$

Hence,

$\red{\bf :\longmapsto\:A= \bigg[ \begin{matrix}2& -1 \\ 1&3 \end{matrix} \bigg]}$

and

$\red{\bf :\longmapsto\: B = \bigg[ \begin{matrix} - 1& -2 \\ 0&1 \end{matrix} \bigg]}$

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.