Question

please the answer . it is urgent​

Answer 1

Solution

We have

$\sf\implies2A+B=\left[\begin{array}{cc}3&-4\\2&7\end{array}\right]$          (i)

$\sf\implies A-2B=\left[\begin{array}{cc}4&3\\1&1\end{array}\right]$            (ii)

To Eliminate A we can subtract (ii)eq from (i)eq

$\sf\implies2A+B=\left[\begin{array}{cc}3&-4\\2&7\end{array}\right]\times 1$       (i)

$\sf\implies A-2B=\left[\begin{array}{cc}4&3\\1&1\end{array}\right]\times 2$         (ii)

$\sf\implies 2A-4B=2 \left[\begin{array}{cc}4&3\\1&1\end{array}\right]$           (ii)

subtract (ii)eq from (i)eq

$\sf\implies2A+B-(2A-4B)=\left[\begin{array}{cc}3&-4\\2&7\end{array}\right]-\left[\begin{array}{cc}4&3\\1&1\end{array}\right]$

$\sf\implies 5B=\left[\begin{array}{cc}3-4& -4-3\\2-1 & 7-1\end{array}\right]$

$\sf\implies 5B=\left[\begin{array}{cc}-1 &-7 \\1 &6 \end{array}\right]$

$\sf\implies B=\left[\begin{array}{cc}\dfrac{-1}{5} &\dfrac{-7}{5} \\\\\dfrac{1}{5} &\dfrac{6}{5} \end{array}\right]$

Now put the value of B on (i)st equation

$\sf\implies2A+\left[\begin{array}{cc}\dfrac{-1}{5} &\dfrac{-7}{5} \\\\\dfrac{1}{5} &\dfrac{6}{5} \end{array}\right] = \left[\begin{array}{cc}3&-4\\2&7\end{array}\right]$

$\sf\implies2A=\left[\begin{array}{cc}3&-4\\2&7\end{array}\right] - \left[\begin{array}{cc}\dfrac{-1}{5} &\dfrac{-7}{5} \\\\\dfrac{1}{5} &\dfrac{6}{5} \end{array}\right]$

$\sf\implies2A = \left[\begin{array}{cc}3+\dfrac{1}{5} &-4+\dfrac{7}{5} \\\\ 2-\dfrac{1}{5} &7-\dfrac{6}{5} \end{array}\right]$

$\sf\implies2A = \left[\begin{array}{cc}\dfrac{9}{5} &\dfrac{-13}{5} \\\\ \dfrac{9}{5} &\dfrac{29}{5} \end{array}\right]$

$\sf\implies A = \dfrac{1}{2} \left[\begin{array}{cc}\dfrac{9}{5} &\dfrac{-13}{5} \\\\ \dfrac{9}{5} & \dfrac{29}{5} \end{array}\right]$

$\sf\implies A = \left[\begin{array}{cc}\dfrac{9}{10} &\dfrac{-13}{10} \\\\ \dfrac{9}{10} &\dfrac{29}{10} \end{array}\right]$

So we get value of A and B

$\sf\implies A = \left[\begin{array}{cc}\dfrac{9}{10} &\dfrac{-13}{10} \\\\ \dfrac{9}{10} &\dfrac{29}{10} \end{array}\right] , B=\left[\begin{array}{cc}\dfrac{-1}{5} &\dfrac{-7}{5} \\\\\dfrac{1}{5} &\dfrac{6}{5} \end{array}\right]$

Answer 2

Step-by-step explanation:

chalo m deti hu

my name is Ashpreet in class 9th