Question

alguem me ajuda a fazer esse calculo? ​

Answer 1

Resposta:

Explicação passo-a-passo:

$\dfrac{X-A}{2}=\dfrac{B+X}{3}+C$

Inicialmente iremos isolar X

$\dfrac{X-A}{2}=\dfrac{B+X}{3}+C\\\\mmc(2,3)=6\\\\3(X-A)=2(B+X)+6C\\\\3X-3A=2B+2X+6C\\\\3X-2X=3A+2B+6C\\\\\\A=\left[\begin{array}{cc}2&1\\3&-1\end{array}\right]\\\\\\3A=\left[\begin{array}{cc}3.2&3.1\\3.3&3.(-1)\end{array}\right]\\\\3A=\left[\begin{array}{cc}6&3\\9&-3\end{array}\right]\\\\\\B=\left[\begin{array}{cc}-1&2\\1&0\end{array}\right]\\\\\\2B=\left[\begin{array}{cc}2.(-1)&2.2\\2.1&2.0\end{array}\right]\\\\\\2B=\left[\begin{array}{cc}-2&4\\2&0\end{array}\right]$

$C=\left[\begin{array}{cc}4&-1\\2&0\end{array}\right]\\\\\\6C=\left[\begin{array}{cc}6.4&6.(-1)\\6.2&6.0\end{array}\right]\\\\\\6C=\left[\begin{array}{cc}24&-6\\12&0\end{array}\right]\\\\Substituindo\\\\\\X=\left[\begin{array}{cc}6&3\\9&-3\end{array}\right]+\left[\begin{array}{cc}-2&4\\2&0\end{array}\right]+\left[\begin{array}{cc}24&-6\\12&0\end{array}\right]\\\\\\X=\left[\begin{array}{cc}6-2+24&3+4-6\\9+2+12&-3+0+0\end{array}\right]\\\\\\X=\left[\begin{array}{cc}28&1\\23&-3\end{array}\right]$